SAT Math: The College Panda [PDF]

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College Panda The College

SAT SAT Math Advanced and Workbook Advanced Guide Guide and Workbook 2nd 2nd Edition Edition

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Introduction Introduction The best best way to do do well well on on any any test test is is to to be be experienced experienced with material. Nowhere is this this more true than than on on The way to with the the material. Nowhere is more true the SAT, which which is standardized standardized to to repeat repeat the the same same question question types and again. again. The The purpose this book book is the SAT, types again again and purpose of this to teach teach you you the the concepts concepts and and battle-tested battle‐tested approaches approaches you questions types types.. If it's you need need to know know for all these these questions If it's n o t in in this this book, book, it's it’s not not on on the the test. test. The The goal goal is is for every every SAT question be a simple reflex, you not question to be a simple reflex, something something you know how to to handle handle instinctively instinctively because because you've you’ve seen seen it so times before. before. know how so many many times

won’t find any any cheap cheap tricks tricks in in this this book, book, simply simply because because there there aren’t that work work consistently. consistently. Don't Don’t buy You won't aren't any any that buy into the the idea idea that that you you can can improve improve your your score score significantly significantly without work.. into without hard hard work

Format Format of the Test There math sections be done minutes without without aa There are are two t w o math sections on on the the SAT. The The first contains contains 20 questions questions to be done in 25 25 minutes calculator. and a a calculator calculator. The The second second contains contains 38 38 questions questions to be be done done in 55 55 minutes minutes and calculator is permitted permitted.. Some made sure accurately divide the practice practice questions Some topics topics only only show show up up in the the calculator calculator section. section. I've I’ve made sure to accurately divide the questions into into non-calculator non-calculator and and calculator calculator components. components.

How How to Read Read this this Book Book For beginning to end. chapter was was For aa complete complete understanding, understanding, this this book book is best best read read from from beginning end. That That being being said, said, each each chapter written possible. After all, you may already already be written to be be independent independent of the others others as as much much as aspossible. After all, you may be proficient proficient in some some topics yet weak others. If so, feel free to on the the chapters chapters that are most most relevant topics weak in others. to jump jump around, around, focusing focusing on that are relevant to your your improvement. improvement. All chapters won't master master the the material material until think through through the chapters come come with with exercises. exercises. Do them them.. You won’t until you you think the questions yourself. If you get stuck on a question, give yourself a few minutes to figure it out. If you’re que stions yourself. you stuck on a question, give yourself a minutes figure out. If you're still stuck, stuck, then then look look at at the solution and and take take the the time time to fully understand Then circle circle the the question number still the solution understand it. Then question number or make make aa note note of it somewhere somewhere so so that that you you can can redo redo the the question Revisiting questions missed is question later. later. Revisiting questions you’ve you've missed the best best way way to improve improve your your score. score.

About About the the Author Nielson Phu Phu graduated graduated from New New York University, University, where where he he studied science. He He has obtained perfect perfect Nielson studied actuarial actuarial science. has obtained scores on the SAT and and on the the SAT math math subject subject test. test. As aa teacher, he has has helped helped hundreds students scores teacher, he hundreds of students throughout Boston Boston and and Hong Hong Kong Kong perform perform better better on on standardized standardized tests. Although he he continues pursue throughout tests. Although continues to pursue his interests interests in education, education, he n o w an an engineer engineer in the the Boston Boston area. his he is now area.

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Table of Contents Contents 1

Exponents Exponents & Radicals Radicals

7

Laws Laws of exponents exponents Evalua ting expressions nen ts Evaluating expressions with with expo exponents Solving equations with exponents Solving equations with exponents Simplifying Simplifying square square roots roots

2

15 15

Percent Percent Percent Percent change change Compound Compound interest interest Percent problems Percent word word problems

33

Exponential Exponential vs vs.. Linear Linear Growth Growth

23

Linear growth and and decay decay Linear growth Exponentia Exponentiall growth growth and and decay decay Positive tion Positive and and negative negative associa association

4

Rates Rates

32

Conversion factors factors Conversion

5 6 6

Ratio Ratio & Proportion Proportion

38 38 44 44

Expressions Expressions Combining like terms terms Combining like Expansion and factoring Expans ion and factoring Combining, dividing dividing,, and and splitting splitting fractions fractions Combining,

7 8

Constructing Models Models Constructing

52

Manipulating & & Solving Solving Equations Equations Manipulating

57

Common mistakes mistakes to avoid avoid Common isolating variables variables Tools for isolating Strategies for solving solving complicated complicated equations equations Strategies

9

More Equation Equation Solving Solving Strategies Strategies More

71

Matching coefficients coefficients Matching lnfinitely many many solutions solutions Infinitely solutions No soluti ons Clearing denominators denominators Clearing

10 10

Systems of of Equations Equations Systems

78

Substitution Substitution Elimination Elimination Systems with with no solutions solutions and and infinite infinite solutions solutions Systems problems Word problems More comp complex systems More lex systems Graphs of systems systems of equations equations Graphs

4

.r

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11

Inequalities Inequalities

91

inequalities solve inequalities H o w to solve How problems word problems Inequality Inequality word inequalities Graphs Graphs of inequalities

12 13 13 14 14

Problems Word Problems

100

Problems & Maximum Minimum M i n i m u m 8: Maximum Word Problems

109

Lines Lines

117

y-intercept and y-intercept Slope and Slope point-slope form and point-slope form and lines: slope-intercept Equations of lines: Equations slope-intercept form form intersection of two the intersection Finding the Finding t w o lines lines lines perpendicular lines and perpendicular Parallel and Parallel vertica l lines and vertical Horizontal and Horizontal lines

15 15 16

126 132 132

Linear Models Interpreting Models Interpreting Linear

Functions Functions function? a function? What is a What undefined? function undefined? When is a function When functions Composite functions Composite Finding solutions to a function a function the solutions Finding the graphs function graphs Identifying function Identifying transformations Function transformations Function

17

146 146

Quadratics Quadratics Tactics for finding roots finding the roots Completing the square square the Completing The vertex form vertex form and vertex ver tex and The discriminant The discriminant Quadratic models Quadratic models

18 18

Synthetic Division Synthetic Division

161

Performing synthetic division division Performing synthetic Equivalent expressions expressions Equivalent The remainder theorem remainder theorem

19

20

21 21

Complex Numbers Complex Numbers Absolute Value Absolute Value Angles Angles

170 170 174 180 180

Exterior theorem angle theorem Exterior angle Parallel lines Parallel lines Polygons Polygons

5

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22

Triangles

187

Isosceles and and equilateral equi lateral triangles lsosceles triangles Right triangles Right triangles Specia l right Special right triangles triangles Similar Similar triangles triangles Parallel Parallel Lines and Proportionality and Proportionality Radians Radians

23

Circles Circles

207

Area and and circumference circumference Arc length length Area of a sector sector Centra l and and inscribed inscribed angles Central ang les Equa tion s of circles Equations

24

Trigonometry Trigonometry

216

and tangent tangent Sine, cosine, cosine, and Trigonometric identities Trigonometric identities Evaluating trigonometric Evaluating trigonometric expressions expressions

25 26 27

Reading Data Reading Data Probability Probability

225 234 244

Statistics Statistics II Mean, and mode Mean, median, median, and mode Range standard Range and and standard deviation deviation Histograms Histograms and and dot dot plots plots problems involving involving averages averages Word problems Boxplots Boxplots

28

Statistics II II Statistics

254

Statistical sampling Statistica l sampling Using and and interpreting interpreting the the line line of best fit Using Margin error Margin of error Confidence intervals Confidence intervals Experimental Experimental design design and and conclusions conclusions

29

Volume Volume

267

30

Answers Answers to the Exercises Exercises

272

6

Ex ponents & Ra Exponents Radicals dicals Here are the Here are the laws laws of exponents exponents you you should should know: know:

Example Example

Law Law 1 xx1z = Xx F

1 331z = 33

,

L

\U_ XO =

11

A

30_ 3D= 11 1

m 0n 111In I 1 111 11 xX",

34 35 Z 39 34. 35 = 39

· x" = x +

x111 if: :

_=

7 27 Z 34 34 37 =

xm‐n

XIII -II

x"

33 4

(32)-l : 38 38 (32)4 =

(XVII) Z _mn 111 (x )11= x 11111

(2 , 3)3 Z 23 3 .33 3

my)!” = : xlllylll xmym (xy)"' I

"I

(2 · 3) 3

Xm

(~)m (y) =z ;:: w

2

,

'1 W1 7 xA_,-mIn = xm

7

3

·3

23

(~)3 (g) =z ~: 3‐3

33 -‐44 1

=2

1

_ g _ 2_ =

34

CHAPTER 1 EXPONENTS & RADICALS

Many students don 't know the difference between

( - 3) 2 and - 32 Order of operations (PEMDAS) dictates that parentheses take precedence . So,

( - 3) 2

= (-

3) · (- 3)

=9

Without parentheses, exponents take precedence:

- 32 = - 3 · 3 = - 9 The negative is not applied until the exponent operation is carried through. Make sure you understand this so you don't make this common mistake . Sometimes, the result turns out to be the same, as in : ( - 2) 3 and - 23 Make sure you see why they yield the same result.

EXERCISE1: Evaluate WITHOUT a calculator . Answers for this chapter start on page 272. (- 1)4 1. l. t‐1)4

3)3 10. -(‐(‐3)3

50 19. 50

(- 1)5 2 2. (‐1)5

11. -(- 6)2

2 20. 332

( - 1) 10 3. 3. (‐])lo

12. - (- 4)3

3‐22 21. 3-

- 1) 15 4. ((21)15 4.

13. 23 X 32

(- 1)8 5.(‐-1)8 5.

14. ( - 1)4

- 18 6. 6. ‐18

15. (- 2)3 X (- 3 )4

2 24. 772

7.-‐(‐1)8 7. - ( - 1)8

16. 30

7- 2 25. 7‐2

s.(‐3)3 8. ( - 3)3

17. 6- 1

3 103 26. 10

9.-‐33 9. - 33

18. 4- l

27. 1010‐33

(- 1)5

22. 53 53

33 X 22

23. 5‐3 5- 3

X

X

8

PANDA THE COLLEGE COLLEGE PANDA THE

The calculator. The a calculator. use a NOT use exponents. Do NOT positive exponents. contains only answer contains your answer that your so that Simplify so EXERCISE EXERCISE 2: Simplify only positive 272. page on start chapter this for Answers you. for done been done have been first ttwo w o have you. Answers this chapter start page

1. =6x5 6x 5 3x 2 · 2x 3 = 1. 3x2-2x3

i

21. 21,

6,‐{1 6u 4 8112 '8uz

2. 2k‐44 ~4k2 : % · 4k2 = 2. 2k-

12

12.

k

5x4 · 3x -‐22 3. 3” 5x -3x

14. 14. (2x 2)- 3 5. (23r2)’3

2 a

(mnz)2

x22 X7. x -3

1 23. % 23x- 2

I l

x mn

4

3x 15_ W3x4 15. (x-2)2

b-‐ 3 .• 3a6. -_ 3a 3a2b 3a ‐ 5b8

3n7 3117 7- 55 6n 3 7.

24. flm2n3 k- 2 k‐z 25. 25- pk-3

3

x1 xi?

16. 7 I

xi x7

(a2b3)2 8. (azb3)2

26.

m2 3 (7) c: n:r

27 27.

(

17. 17 x2. x2 x3. ‘r1 x4 x4

W4

9. ( xy4 ) x3y2

9,(”fl/2) ‘

2)- 3 • 2x 3 18. (x (.r2)'3-2x3

3

(- x) 3 10.' -_(_-")

(b- 2) -3. (b3)2 (b‘2)‐3 , (b3)2

)3 (m2n ("1 fl)‘ 22. ‐‑(mn 2)2 22.

- 4u 2v 2u v 2 • ~4uzv 13. 21402-

-3 - 3111 7m 3 •-‐3m‘3 4. 4. 7m3

10

a- 2)2 a- 1 •~a‘2)2 20. ((a‐1

(x2y - 1)3 11. 11. (x2y_1)3

3 )2 ) 2 •- (3111 19. (2m (2m)2 (3m3)2

2a

24 ) x y~z4 x2y3 4z-5 y -3 x x‘3y'42“5

? off y ? terms o the value what is the then what 3x+2 = y, then EXAMPLE If3H2 value ooff 33x j r in terms EXAMPLE 1: If

A )y+9 A)y

B ) y-- 9 B)y

ql C)% 3

D)%

only difference the exponent the 2 in the that the notice that we notice Here we what x is. Here finding what trouble of finding Let’s exponent is the the only difference the trouble avoid the Let's avoid out: 2 the extract let's exponents, laws our using So want. we what between the given equation and what we want. 50 using o u r laws of exponents, let’s extract the out: and equation given the between

3X+2:3X.32=y = 3x . 32 = .1/ 3x+2 3xx=: 'i 3 9

Answer Answer ~(D) . a?7 value of a the value what is the ra +7 , what = 3‐“7, 3a+l = ff 3"+1 EXAMPLE 2: If EXAMPLEz-

equal. be equal. therefore be muu sstt therefore the same. are the bases are the bases Here same. The The exponents exponents m that the Here we see that

a + 1l ==-~ aa+ +7 a+ 2 2aa=: 6

au=@J =l

9

CHAPTER CHAPTER 1 EXPONENTS EXPONENTS & RADICALS RADICALS

a 4a EXAMPLE 3: U 2a - b = 4, what is the value of b ?? EXAMPLE3zlf2a‐b=4,whatisthevalueof§5

2

2. Realize Realize that that 4 is just just 222.

(22V 22" 4 2. ?:_2h_=F:2 :2

4"

Za‐b

Square Square roots roots are are just just fractional fractional exponents: exponents: I

.Jx

xi =

x?l But what i ? The top means bottom means means to cube cube root root it: what about about xxi? The 2 on on top means to square square x. The The 3 on the bottom 3/x2

We this more Wecan can see this m o r e clearly clearly if we we break break it down down:: g

l

2 2 I 3;-;, 2 (x (x2)? = {7x2 xx33 = )1 = v x2

The order -rooting doesn’t doe sn't matter order in which which we we do do the the squaring squaring and and the the cube cube‐rooting matter.. 2

x? = (xhz : (3/32 The result just the cube the outside. outside . That That way, we don't need need the the parentheses. parentheses. The end end result just looks looks prettier prettier with with the cube root root on on the we don’t

EXAMPLE EXAMPLE 4: Which Which of the the following following is is equal equal to ~ Vac‐5 ? ? A) A)x

B) .x5- x4 B)x5‐x4

CM;

on?

1 The fourth root fractional exponent The fourth root equates equates to aa fractional exponent of 3', so so

4

4

x5=x

Answer §). (C) . Answer

10 10

PANDA COLLEGE PANDA THE COLLEGE THE

root, factor square root, a square "surds"). To simplify called ”surds"). square roots simplifying square on simplifying you on also test you SAT will also The SAT roots (also called simplify a pairs : any pairs: out any and take out root and square root inside the square number inside the number the

JT:\/2~2.2~z= \J /[I1]-[I}] - - ' 3 = 2 --32=\2·2v3 / § = 4= \4v3 /§ v'48 = ✓2-2·2 · 2 · 3 = [I}] . pair -. second pair out for the second another 2 out Then we take another [I}] . Then the first -. out for the a 22 out take a we take above, we example above, In the example looked have would route quicker a course, Of 4. get to root square the outside 2's two the multiply we multiply Finally, we t w o 2’s outside square root course, a quicker route would have looked this : like this:

\/4‐ = t✓~/ _ - 3 =: 4\/§ 4v3 v'48 = example: another example: Here's another Here's

M: \/--~2=2-3\/§:6\/§ pair: as a pair: root asa square root under the square back under put itit back and put outside and number outside the number take the backwards, take To go backwards, To

N \/6-6- = V72 = x/fi 6v'2i =: ~ any triplets out any ifi6, take as W, cube root such a cube simplify a To simplify such as take out triplets::

W=x7-2=2¢/§ 0? where x > O? ¾, (x2 ) Q form of (12) equivalent form the following Which of the EXAMPLE 5: Which following is an anequivalent , where A)..jx AWE

CW?

x..fi B) Bm/E

DN/E

Solution 1: Solution 1: 3

s

1

(x2)“ zxz4 ZYZ

zmz

--x=x\/§

Answer Answer §)(B) . 3

. . 73 23 2 . exponent of 3 to the exponent of x in , each of the compare this (x2 ) 4 z= x 44 = Solution 2: Since (x‘)El Solution z x 2- ,, we we can can compare exponent 5 exponent each the 2,1

3

3

2

answer choices. choices. answer I

,Ix== xx2 Choice Choice A:A: J?

1 I

1 I- iI

5 3

ChoiceB: Choice 8:

= xx21 x\/§=x‘-xz = x 1 3 x,/x = x1 • x 2 =x”Z

Choice Choice C:

W

Choice Choice D:

é/E : xxI1 -1/x=

2

= x‘1 1 I

that the answer confirm that results confirm These answer is B. These results

11

& RADICALS CHAPTER 1 EXPONENTS & RADICALS

EXERCISE 3: Simplify the radicals or solve for x. Do NOT use a calculator. Answers for this chapter start on

page 272. page272.

1. \/1‐2 l. Ju

/128 ' 10. v128

2. 2. fi)%

11 ll.Sv'2 = ./x

3. v'45 1/5 3.

12. 12 3,/x

4. \/1_8 /18

13 2v'2= v'4x 13.

5. 5 2/fl 2m

14. 144)6

6 3\/7_5 6. 3/75

15

7. «3‐2 7.J32

16. 16 4/3x

8. m v'200

17. 17 3v8

= x/2

9. 9. x/é v8

18. 18 x,/x

= v'2l6

12 12

= v'45

= 2/3x

= 2,/6

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CHAPTER EXERCISE:Answers for this chapter start on page 272.

A calculator should NOT be used should N O T be used on the following following questions. questions.

If

__ 1

~ xc for all positive positive values what is \/x‐ = x‘ values of xx,, what V} 3

yX

the value value of c ?? the

I

If aa- 1f = what is the = 3, 3, what the value value of a? a?

A) A) - 9 1 B) 9 1 C) 3

D) D) 9 If 3x 3x = = 10, what is the lu e of 3xIf 10, what the va value 3"’33 ?? 10

A)

~ 3

A ) yy ++ 3 A)

B)

39

B)) yy-‐ 3 B C) C ) 33- ‐ yy D) 3y 3y

10 10 C) 57 27

2xx

IfIf 27 Y = 2

3, then then x must must equal equal 223,

10 10

27 27 D) fi 10

10, what what is the the value value of y2 y °? If y3/55 = 10,

IfIf a are positive integers, which a and and b bare positive even even integers, which of the the following following is is greatest? greates t?

A ) 440 0 A) B) 400 B) 400

A) A) 8) B) C) C)

C) 1,000 1,000

D) 10,000

(- 2a)b (‐2a)" ( - 2a) 2b < x . x h = miles 1 hour 2 36 minutes 1es x our 50 m1

Leona will take take Leona will

x 30 miles miles>
1, 1, which which of the the following following is equivalent equivalent to

? ‐‐1‐xT1 ? 1 X

x --‐ 1 + x + 1 2x22 2x __ A) (x - l)(x + 1)

A)(x‐1)(x+1)

B)

22 _______

B’(x‐1)(x+1) (x - l)(x + l)

C)

x(x‐1)(x+1) x(x - l)(x + 1) _______ 2 2

C)

D __

)(x+1) D) ((xx-‐ ll)(x+l) 2

)

2

First, combine fractions on the common denominator denominator (x -‐ 1)(x + First, combine the the two t w o fractions the bottom bottom with with the the common + 1). 11 1l X _ x +1 -x-- 1 + -x +-1 = -(x--- 1)-(x-+~1)

X x -‐

1l

2Xx 2

x--- 1-)-(x_+_l_) = -( x‐1+x+1“'u‐1nx+u + -(u‐1Mx+n ux--‐ 11-)-( x xx_+_ +l_n)

Next, Next, substitute substitute this back back in and and flip it.

+11)) x _ x (x ( x-- 1l ))(x ( x + 11)) _ ((xx -‐ 1l ))(x (x+ X ‐ ‐ 22xx ‐ _ _ ‐ ‐ 22x§ _ _ ‘ ‐ ‐ 22 _ ‑ ( x -‐ 11)(x ) ( x + 1l ) (x Answer ~(D) . Answer

47

CHAPTER 6 EXPRESSIONS CHAPTER EXPRESSIONS

5. Splitting Splitting fractions fractions 5. 30

EXAMPLE 7: Which EXAMPLE Which of the followingisequivalent to 306+ the following is equivalent to : C7 c?

+c A) A S5) +c ? 6

10+c B) 10 +c B)‐2‐‐ 2

C ) 5 +c +c C)S

We can split Wecan split the the fraction fraction into into two: two :

ic

D)5+g 0)5 +

30+c_?1)+£_5+£ 6

_ 6

6 _

6

The answer is ~( D ) .- This The answer just the the reverse adding fractions. fractions. This is just reverse of adding Note that Note that while while you you can can split split up the the numerators fractions, you so with numerators of fractions, you cannot cannot do so with denominators. denominators . So, 3

3

x +y

X

_ 7g_ -i=-+x+y x

3 _ y

3 !J1 you cannot cannot break break up a fraction fraction like In fact, you like -x i -y any further.. any further

x +y

48

COLLEGE PANDA THE COLLEGE PANDA

CHAPTER EXERCISE:Answers for this chapter start on page 284.

should N A calculator should O T be NOT be used used on the following questions. following questions.

4 + 8x Which of the Which the following following is equivalent equivalent to to 4 ~:x 12x forxX76 for /; O 0??

Which of the the following following is equivalent Which equivalent to 2y + 6x23] + 6ch2 6x 6xy 2 ?

1 ++ Bx 8x 3x 4 + 2x B) B) 3x 1 1 + 22xx C) 3x

A) A)

A) A) 6xy(x 6xy (x + y) B) 12xy(x + + y) 2 6x y2(y + x) C) 6x2y2(y 3y 3 D) 12x 12x3y3

D D)) 1l

0, then 2 !+ ~ is equivalent equivalent to which lIff a ,f; 76O,then + 2 which of the the a

Which of the the following following is equivalent Which 3x44 -‐ 3 3 ?? equivalent to to 3x

4

following? following?

A) 3(x 2 + 1)2 A)Mx2+1V

3 + 4a 4a 4a 4 ++33a a B) 4a

B) 3(x2 1)22 3(x2 - 1)

A) A)

C)

D)

C) C) D) D)

3 ( x 3 -‐ 30:

1”(35+ ) (X + 1) 1)

3(x2 +1)(x+1)(x + l )(x + l )(x -‐ 1) 3(X 1)

7

4a 4

(x +1)2 +2(x +1)(y 1)+ ( y ++ 1) 1)22 + 1)2 + 2(x + l )(y + + 1) + (y

a+4

Which of the Which the following following is equivalent equivalent to to the the expression shown expression shown above? above?

A) (x ++ yy ++1)2 A) (x 1)2 Which of the following is equivalent to (x 2 + y) (y + z) ? A) x 2z + 1/

+ 2)2 B) (x (x + +y + 2)2 C)) ( (x x ++yy)2 F ++22

+ 1/Z

D) (x + y)2 y )2 -‐ x -‐ y

B) x 2y + x 2z + y 2 + yz C) x2y + y2 + x2z D) x2 + x 2 z + y2 + yz

49

CHAPTER 6 EXPRESSIONS EXPRESSIONS CHAPTER

If yé0 yéy, If y I0 and and xx Iy, which which of the the following following is is 2 xy - x equivalent to xy ‐ x;2 ? equivalent ?

A)‘i‘ A)

The expression 8x 2 - ~y 2 can be written in the

form where c is aapositive form 8(x -~ cy)(x cy)(x + + cy), where positive

xxyy -" yy

constant. What What is the constant. value of c ? the value

_ 'f_

1 A) A) 16

X

B) 'f_ i X

1 B) 8

xX y

1 C) C)

C) g-

4

xX y

D) D) 7

D)

which of the following If x > 1, which following is equivalent equivalent to

2 xx2(x (X

1 -----,----1 -=?7 x - 1 x +S --+-2 3

x‐1+x+5'

C)

D) D)

2) + +44 + 2)(x 2) ( X -‐ 2)

Which of the the following following is equivalent Which equivalent to the the expression above? expression above? A) (x2 (x2 ‐- 2)2 2)2 A) B) B) (x2 (x 2 + + 2)2 2) 2 C) C) (x (x --1)2(x 1) 2 (x + 2)2 2) 2 D) D) (x (x +1)2(x + 1)2 (x -‐ 2)2 2) 2

A) 5): Sx + + 77 A) B) B)

~

6 6

2x + 4 6

5x + 7 1

30x + 42

1

2+ X -- 1 2- X

If x I76O, equivalent to to 0, which which of the following following is equivalent the given given expression? the expression? 2x -‐ 1 2x + 1 2x + 1 B) B) 2x -‐ 1

A) A)

4x22 -‐ 1 4x x2 x2 D D)) -‐ 1 C)

50

THE THE COLLEGE COLLEGE PANDA PANDA _\

A calculatoris allowed on the following A calcu

I‑

lator is allowed on the following quest ions. questions.

I

‐

“

): X

2

3x + 8x2 8x 2 ‐- 4x 3x33 +

7x2-11x - 7 712‐11x‐7 Which of of the following is the the sum Which the following sum of of the the two two polynomials above? above? polynomials

”

X A) A) _L - x- 2

x‐2 rX

B) 2(x B) -_ T {f -

C)

C) 10x 10x55 -‐ 7x -‐ 7 2 4 + 3x 3 D) D) 1Sx 15x4 + 3x3 -‐ 15x 15x2 -‐ 7

D) 3x D) 2(x - 2) 2(x ‐ 2)

Whic h of ing is equivalent to Which of the the follow following is equivalent to the the expression above? above? expression A) -‐2a\/E 2afa A) B) afa B) a\/E C) ) 3a C 3 -a ‐2/a 2 fi D) D) 3a 3a+8\/E + BJa

9(2y) 2 2 + 2(6y) 26 22 7 If yyéo, y f=O,what what is is the the value If MW · B(3y)2 ?· value of

H

"

Which of the the following Which following is is equivalent equivalent to to the the expression above above for x 752 =/=2 ? expression ?

3 + x 2 - 15x A) 3x 7 A) 3x3+x2‐1Sx‐7 3 B) 3x 1Sx2 -‐ lSx B) 3x3 + +15x2 15x -- 77

(5a + + 3Ja) WE) -‐ (2a + + sJa) s fi )

x

--+~ --+2(2 X- x)a ‐

x- 2

xX

C) 2(x - 2) 2(x ‐ 2) 3x

2) 2)

Models ConstructingModels Constructing graphs . expressions, equations as expressions, quantities as real-life quantities represent real‐life you to represent require you questions require model questions Constructing model Constructing equations,, and and graphs. chapter is specifically focused but this chapter book, but chapter s in this book, other chapters can be type can Questions be found found in several several other Question s of this type Now exponential). Now quadratic , exponential). linear, quadratic, types (e.g. linear, pertain to any of the conventional 't pertain ones that don on on the ones don't conventional model model types be won 't be there won't quite simple, actually quite chapter's difficult. Most mean this chapter’s doe sn't mean that doesn’t Most of the questions questions are actually simple, and and there in the chapter leave the rest to you in and leave example s and two examples We'll just do two here . We’ll concept s here. any nnew e w concepts chapter exercise.

i.

. l

schooll>µys th.e school grade - If each grade. students in each grade levels school, there a school, EXAMPLE 1: At a there are are a grade levels with with b students If the buys n 71 EXAMPLE s ckers thenumber gives the the following which of the equally among stickers bedistributed among the the students, students, which following gives number of stickers distributed equally stickers to be each receives? student receives? each student ab ab A) A) 7 n

an B) B) an 7

b

bn C) bn a

rt

D)~El? D)

C) 7

ab

divide student s. To find the number b) = The school has a (a) (b) : ab abstudents. number of stickers each each student student receives, we divide a total of (a)(

ITe:IJ.

~: . Answer student s: ‐”‐. the number Answer (D) . number of students: number of stickers n by the number ab

th drained tank.was until it was rate imtil was pumped pumped into into a tank at a constant constant rate was foll. full. The tank was then EXAMPLE 2: Water was amount of total ent-the rep could graphs at been filled. Which of the the following following could represent the total amount had been than it had rate than slower rate a slower at a water the tank time? versus time? tank versus water in the A)

B)

C)

.... s ....2

0 s....fil § .... gt

~

s.

§2 as

at ~~

s“ ~

s;

i” l

~~

0~

Time Trme

D)

..... 0

“6 ...Iii

0

l‐i

533

~j

s1

:

Time Time

Time Jhne

Water be represented and to to the right (positive slope) . (positive slope). the right up and going up line going a line by a represented by should be tank should the tank into the pu~ped into be~ng pumped Water being Water slope).. That That leaves leaves (negative slope) right (negative the right to the and to down and going down line going by aa line represented by be represented should be dramed should bemg drained Water being us answers C and D. Since tank was drained at at aa slower slower rate is ~(D) answer is the answer filled, the was filled, than itit was rate than was drained the tank Since the and D. wi~ answers~ ~s with

line o t as steep as as the going up up.. line going the line as steep 1snnot down is going down lme going

52

\

I

i

~it

0~

Time Tlllle

5;

the the

THE COLLEGE THE COLLEGE PANDA PANDA

CHAPTEREXERCISE:Answers for this chapter start on page 285.

should N NOT A calculator should O T be be used used on the following questions. following questions.

An intemet charges internet service service provider provider charg es a a one one time time setup setup fee of $100 and and $50 each each month month for for service. service. If c customers customers join at the the same same time time and and are are on the of the the following following the service service for formm months, months, which which of

A carpenter carpenter lays lays x bricks bricks per per hour hour for y hours hours and then and g bricks then lays lays ~ bricks per per hour hour for 2y more more

expressions the total total amount, amount, in expressions represents represents the dollars, provider has charged these these dollars, the the provider has charged customers? customers?

2

hours. terms of x and hours. In terms and y, how how many many bricks bricks did did he lay in total? he total?

A) 100C A) 100c + 50m

A) A) 23:31 2xy

B) 100c 100i: + 50cm

5

B) B) gxy xy

C) 150cm

C) 5xy Sxy

D) 1 100m + 50cm 00m +

2

3 D) D) gx+3y 2x + 3y

At aa math team competition, there are are m m schools math team competition, there schools with from each school. The The host host with n students students from each school. school wants order enough such school wan ts to order enough pizza pizza suc h that that there student. If there are there are are 2 slices for each student. are 8 slices in one which the following following gives one pizza, pizza, whic h of the gives the schooll must the number number of pizzas pizzas the the host host schoo must order? order? mn nm A) 8 mn B)

A cheese vendo r current ly has 175 pounds of mozzarella avai lable for sale . If each pound of mozzarella sells for $8.75, which of the following functions gives the amount of mozzarella M, in pounds, still available for sale after d dollars worth has been sold?

d M (d) = 175 - 8.75 B) B) M(d) M (d) =175‐8.75d = 175 - 8.75d

A)

T4

C)

8.75 C ) M(d) 7 5-‐ -8 ? C) M (d) = 1175 d D) =175(8.75)‐d D) M(d) M (d) = 175(8.75) - d

m+2n m + 2n 8 8

D) D) 2mn

A retail store has monthly fixed costs of $3,000 and monthly salary costs of $2,500 for each emp loyee . If the store hires x emp loyees for an entire year, which of the following equations represents the store's total cost c, in dollars, for the year? A) A) cc == 3,000+2,500x 3,000 + 2, 500x

B) B) c c==12(3,000 12(3,000 + 2,500x) 2, 500x) C) cc = 12(3, 12(3,000) + 2,500x D) cc== 3,000 3,000 + 12(2,500x) 12(2,500x )

53

CONSTRUCTING MODELS CHAPTER MODELS CHAPTER 7 CONSTRUCTING

A calculator is allowed on the following questions.

A manufacturing plant increases the temp erature of a chemical compo und by d degrees Celsius eve ry m minut es. If the compound has an initial temperature oft degrees Celsius, which of the following expressio ns gives its temperature after x minutes, in degrees Celsius?

A) A) 8) 3)

biking for 4 commu te by biking began a 5-mile cormnute Kaiba began rest area the rest at the stopped at area. He stopped rest area. a rest miles to a walked for the then walked and then for 15minutes 15 minutes and faster bikes faster If Kaiba bikes commu te. If remainder of the commute. remainder grap hs following graphs which of the following wa lks, which he walks, than he than commu te? his commute? represent his cou ld represent could

m x++ t mx d m mdd++ t

A) A)

xX

"U -0

QJ

d C) t +C) t+‐d‑ mx dx D) Hi" D) t +m

Q) o:i 2

E: 4

2 Ez w s.is 11 u

C: -

6"' -‐ 0 D (I)

0-L---+---+----+-➔

45 30 15 15 30 45 (minutes) Time (minutes)

8) B)

bakers to make emp loys bakers cupcake store employs A cupcake make boxes and cupcakes x tains con box Each cupcakes. Each of cupcakes. contains cupcakes and cupcakes y produce to expected baker is expected each baker produce cupcakes expressions following expressions Which of the following each day. Which each needed for all boxes needed number of boxes gives all the the gives the number working for 4 bakers working produced by 3x bakers cupcakes cupcakes produced days? days?

T: -0

QJ

o:i Q) 2

5.

~ ,...._4 . E? 4 [r= 2_ 3 m- ~ 3 2 u E 2 a5 11 s2 QJ ·-

C: -

(I)

6"'

'‐

D

A) 12x2y 12x2y 3 3y B)7y 8) 4

C)

5

~ ,...._4 F r:~ 33 . QJ · -

0 -"""'--+---+----

o

15 30 45 Time(minutes) Time (minu tes)

C) C)

12x2

‘U -0

QJ

y

0.0 o:i 2

5

4 ~ ,...._4 «EA r= P g~ 33

D) 12y

·;: éé z211 s2 ~ C: ..::,

U)

-6"'

5

o 0

souvenir price of one At shop for tourists, one souvenir touri sts, the price At a shop purchased additional souvenir dollars . Each additional is a dollars. souvenir purchased Which percent. 40 percent. Which discounted by 40 after after the first is discounted C, cost C, total cost gives the total equations gives of the following following equations where souvenirs, where purcha sing n souvenirs, dollars , of purchasing in dollars, n> > 11??

15 30 45 30 Time (minutes) (minutes)

D) D) -0

QJ

o:i

>

0, 0, the the answer answer is is I]. I. Because the tes that

Be comfortable comfortable solving solving for expressions, expressions, rather than any variable 7. Be any one one variable EXAMPLE EXAMPLE 10: If 3x 3x + + 9}; 9, wha whatt is is the value of x + + By? 9y = 9, the value 3y ? Get in the the habit habit of looking looking for what what you you want want before before you solve for anything to get get you solve anything specific. specific. Is Is there there any any way way to the answer the answer withou withoutt solving solving for x and and y? Yes! Dividing Dividing both sides of of the the given given equation equation by gives x + Yes! both sides by 33 gives + By: 3y = I. [I].

EXAMPLE EXAMPLE 11: If ~ 5= = 3, is the the value value of iY ?? 3, what what is 2X y 1 A) E

1 B) 5

z C) ~ C) 3

~3 D) 5 2

D)

Here, Here, we we have have no no choice choice but but to so solve the express expression. but we lve for the ion . We’re We're given given x over over y but we want want y over over x. We We can can flip the given equation get the given equation to get I/ 1

=-

!,.._

3

X

ix

Then we can divide both both sides obtain Then we can divide side s by by 22 to to ob tain the % we’re we're looking looking for: for:

63 63

ix=

21x

1.

1 1 = 8“ Answer Answer (A) . = 2 3

: ‐3~

[EI].

CHAPTER 8 MANIPULATING MANIPULATING & SOLVING CHAPTER SOLVING EQUATIONS EQUATIONS

8. In In some some cases, cases, you may need need to plug answer choices guess and and check check you may plug in the answer choices or guess When you When you can’t ”mathematical” way way to get get the the answer, answer, you you have w o options: options: 1) plug plug in the answer can't find a a "mathematical" have ttwo the answer choices or 2) guess choices Both are are valid valid strategies strategies that that you shouldn't be afraid to use. Not only does aa guess and and check. check. Both you shouldn't be afraid use. Not only does "brute force” ”brute often tum turn out out to be be quite quite efficient, efficient, but, but, for some some questions, questions, it is the the only only way way to get get the force" approach approach often the answer . answer. EXAMPLE EXAMPLEIZ: 12:

v22~x=x‐2

What the set set of all solutions solutions to the the equation What is the equation above? above?

A) {- 3, 13} A){‐3’13}

B){3,6} B){‐3/6}

C) {13} C){13}

D) {6} D){6}

We could could solve solve for x by We by squaring squaring both both sides, sides, but but plugging plugging in the the values values from choices is actually actually from the answer answer choices much quicker much quicker and and easier. easier. We We just just have have to see see which which of the the values ( ‐ 3 , 6, or 13) satisfy the equation. When values (- 3, satisfy the equation. When x= the left hand side = -‐ 33,, the left hand side is y'25 m = z 5 and and the right hand hand side z -‐ 55,, so so -‐33 is not in the the right side is -‐33 ‐- 2 = the solution solution set. set. When x =z 6, hand side /16 == 4 and When 6, the the left left hand side is m and the the right right hand hand side the solution side is 6 -‐ 2 == 4, so so 6 is in the solution set. set. At this point, point, we can answer is~ this can tell the the answer is m

based on the the available available choices, u t let’s 13 just be sure. sure. based on choices, b but let's test test x = = 13 just to be

When x = When z 13, the the left hand \/§ = : 3 and and the right hand 13‐- 2 = 11, so so 13 13is solution hand side side is ,/9 the right hand side side is 13 is not in the the solution set. set.

EXAMPLE EXAMPLE 13: 2 (x3 -‐- 4) = 4x xx2(x3

If xxisis an an integer, integer, what what is is one one possible possible solution solution to to the the equation equation above? H above?

Assuming we can't can’t use use a a calculator, calculator, there there is no no easy easy way way to solve hand, and and there there are are no no Assuming solve the the given given equation equation by hand, answer like this this calls calls for guess and check. check. Typically, Typically, you you want want to to start start with with answer choices choices to work work from. from. A situation situation like guess and small numbers like : -‐ l1,0, , O,1, small numbers like xx = 1, and and 2.

Il won’t process here here since since it’s obvious what what you you need do. It It turns won't work work through through the the guess guess and and check check process it's obvious need to to do. turns out out the the

answer l. answer is is [II.

64

THE COLLEGE THE COLLEGE PANDA PANDA

EXERCISE1: Isolate the variable in bold . Answers for this chapter start on page 287. l. A= nr 2 A=7rr2

22. If If t = gax, ~ax, find find ax in terms terms of oft.t.

2. C ==22nr 7rr

23. If 3x + 6y = 7z, find x + 2y in terms of z.

fig...

+5 =

3. A = ~bh

24. If x

wh 4.. V V :=I lwh

25 25.. If a;

2b, find 2x

+ 10 in terms

of b.

= a, find 4t in terms of a. ”dz‐fl1 =a,find4tintermsofa.

V=7rr2h

p-h p ‐Iz

2 2. .

If p + h

p

‑ · find 11 6 Ifp+h =z -33, fmdh

2 26..

.V:7rr2h : a¥2 ++ b§2 7.. c82 =

I

. terms o f r. 1 + 2r = -1 , find t m 27 . If -27. 1- t 2

If lltztrzéfindtintermsofr.

8..V=s3 V = s3

S227trh+27tr2 9.. S = 2mh + 2m 2

4x411 4x

a -2 C 2 b =d

If 29. If

C ‘ab1 _ £ d

30. 1f2x(x 3 ‐ 30.1f2-‘(x3

lO. b _ d

ll.

2.1/ terms xY = z, then then find find x z. 28. If x” : 2, xz?’ in terms of 2.

x3 -

X2

4 ), what is pin terms = p(x p(x55 -‐ x14), what p in terms of x? x?

1 -1) = m(x + 11)-: , whatismin 1) = m(x2 )‐ ‐,w h a t i s m in x x2

b _ d

xz‐xl

2

terms of xx?? terms

y ==m 12. y mxx + +bb

"IZ‐y2_y1

2

”M_x3=

whatisnintermsofx? 31. lf Jx+2 l - x3 = 2..., l,whatisnintermsofx? 5x2 -‐ 3 nx x 5 n If a(b2 + what is a in terms terms of of 32. lfa(b2 + 22)) + c = = S(c 5(c + 1)3, l)3,what isain

b and cc?? band

m=y‐2_y1 xz‐x1

33. If If k(x2 k(x 2 + 4) + + 4) + ky = z

2 = “22 + 15. v02 = 11 + 2as

a

16.

xandy? x and y?

X

bE =_y2l b_y2

34. If

ax + 3a + x

andb? and b?

t=27r\/Z = 2nJ!j

17. t

7x2 + +33 h . k . 7x2 f , what w at 1s terms of o is k m in terms 22

8

/p+q 18. A z=mm, ✓p+q

I f XX == :X+1,findXintermsonandZ. 19. If : ~, find X in terms of Y and z.

Y+Z

20. If If x(y + + 2) = y, find find yin y in terms terms of x.

~a = 1a 126 ++1] , find terms of “E‑ find a in terms ofbband and c. b 2c

21. If

65 65

+ 3 = b, what

is x in term s of a

CHAPTER 8 MANIPULATING

& SOLVING EQUATIONS

CHAPTER EXERCISE:Answers for this chapter start on page 287.

used on the NOT should N A calculator should O T be used questions. following following questions.

If;=

~m, what is the value of m?

1 A) 6 A) 6

Ifa+b=‐2,then(a+b)3= If a + b = - 2, then ( a + b)3 =

2 B _ B) 3

A)) 4 A

) 3 5 C) 3

B) O 0 C)) -‐ 4 C

6

D)) -‐ 8 D

D) 6

? 4) 2 ? + 4)2 4)2 = (n + 11 is (n ‐- 4)2 va lue of n what value For what For

IIff 3x 3 x++ 11 = = -‐8,whatisthevalueof(x+2)3? 8, what is the va lu e of (x + 2) 3 ? A A)) -‐ 1 B) l1 C) 8

D) 125

! i,

ac?? of b -‐ ac the value what is the value ofb If %x ~ x ~g == 1, what

a

‐k 4+ 2 ‐= ‐3 " ,w here kk i= ¢ ‐- z2,, what w ha t i is 's k ' etterms rms kinm where If k 2

C

o off xX??

A A)) -‐ 3

A) 12‐2x 12 - 2x A)

B) 0

xX

C) 2

B) 12+2x 12 + 2x B)

the from the D) It cannot determined from be determined cannot be given. information given. information

xX xX

C) C) 12+2x 12 + 2x D 1 2 x- ‐ 2 D)) 12x

6x -‐ 7? value of 6x the value If 3 , what 7? what is the = -‐ 223, 3x -‐ 8 _‐_ If 3x A A)) -‐ 5

va lue of the value what is the and x < 0, what If 36 and = 36 3) 2 = If (x ‐- 3)2 x2?? x2

B 1 B)) -‐ 221

C C)) -‐ 3300

D 7 D)) ‐- 337

66

THE COLLEGE PANDA

r-

f=p 0 and > Oand

J

of yy?? value of the value (g)3 what isis the , what = 31/5, (~)3 2

an value J future value the future gives the above gives The f of an formula above The formula the payment p, monthly payment the monthly on the based on annuity based annuity p, the months n. the number and the rate i, and interest rate interest number of months terms of J, pin gives following the Which of the following gives p in terms Which f, i, n?? and n and

fiJi

A) (1 + i)" - 1 M u+nn‐1 ((11++ i)i ) " ‐ 11 B ‐ fi ‐ B) ) Ji 11

-

If

fJ ‐- i

C)

2\/x+4 2Jx + 4 3

. value the value 1s the what is and x > 0, what = = 6 and

off X? o x?

+ i)" - 1 C) (1(1+i)"‐1

D fi + 1 ++ i i) ) "11 D)) Ji +11 ‐- ( (1

.

n1

n

value of -L ?? the value 1s the what is 2, what = 2, If fl- = 2m 211 Zn 2m A)

1 8

2 20‐fi=§fi+1o 20 - ✓x = 3 ✓x + 10

1

B) 4

equation the equation value of xx is the what value If x > 0, for what true? above true? above

1

C) -

2

D) 1

x 2 + Sx - 24

=0

0, < 0, above and equation above the equation solution of the If k is a solution If and k < ? lkl of |k| ? what value the value what is the

the value + y2 + 16, what Jx2 +y2 IIff x +y what is the value of + y = «X2 xxy? ? l/

67

CHAPTER 8 MANIPULATING & SOLVING EQUATIONS

- 1 xx ++2 2

2X

X -

A calculator is allowed on the following questions.

2

_4 T

What is the the solution the equation equation above? What solution set set to the above?

y+2kx=kx2+5 y + 2kx = kx 2 + 5

A) A) {{‐ 10,0} 10,O}

In the the equation equation above, above, k is a constant. constant. If If y := 23 what is the the va lue of k? when x = 3, what when value k?

B) 10, -‐ 44}} B) {{‐10, C) C)

{0,8} {0,8}

A)) -‐ 6 A

D)) {D { 44,8} ,8}

B) 3 C) 6

D) 9

(92 -‐2(%) 15==o, what isthe 2(i) -‐ 15 0,whatisthe

IIff xx > > 0and Oand (if value value of xx??

xX = _ x+12 6 X + 12 .' ,what 1s the value of -_ If -_ 6 =-42 ,whatis the value of x ?7 . ~

6

X

1 A) 3

3

B) ) 2 B C) 3 C) D) ) 6 D 2

x - 4x

+3 = 4

x- 1

What above? What is is the the solution solution to to the the equation equation above? d

= a (~)24

Doctors use rule, shown above, to Doctors use Cowling's Cowling's rule, shown above, to determine right dosage dosage d, in milligrams, determine the the right milligrams, of medication for aa child child based on the the adult adult dosage dosage medication based on a,in child’s a, in milligrams, milligrams, and and the the chi ld 's age age c, c, in years. years . Ben is a who is in need need of a Ben a patient patient who a certain certain medication. uses Cowling’s medication . If a a doctor doctor uses Cowling's rule rule to prescribe prescribe Ben Ben a a dosage dosage that that is half half the the adult adult dosage, Ben's age, dosage , what what is Ben's age, in years? years?

2 4 : 8x 8x44 xx2(x4 (x -‐ 9) =

A) 7

0, for what what real real value value of x is the the equation equation If x > 0, above true? true? above

B) 9 C C)) 11

D 3 D)) 113

68

THE PANDA THE COLLEGE COLLEGE PANDA

111 3‐1 ,______

V --------'Y

---------

Qu estio ns 23-24 refer to the following Questions following

1

-

A) B) C) D)

the figure figure above, In the above, ttwo w o objects objects are are connected connected by aa string which is threaded threaded through string which Using its through a pulley. pulley. Using weight , object object 2 moves moves object object 1 along weight, along a flat surface. surface. The acceleration acceleration a of the the two two objects objects can The can be be determined by the the following determined formula following formula

A) A) B)) B

---'~--'----=

C)) C

D)

where m1 where object 1 and and m, and and m2 are are the the masses masses of object object 2, object 2, respectively, respectively, in kilograms, the kilograms, g is the acceleration due due to Earth's acceleration Earth's gravity gravity measured measured in .

The acceleration acceleration would The would be be quadrupled quadrupled (multipled by (multipled by a a factor factor of 4).

A ---------

2y) ‐- 3z lf 3(x ‐- Zy) 32 := 0, 0, which which of the the following following expresses x in terms terms of y and and z ?? expresses

a _ m2g ng -‐ w11 ”mg 1g a= m + m z Ill]] + /112

m

The acceleration acceleration would The would stay stay the the same. same . The acceleration acceleration would The would be be halved. halved . acceleration would The acceleration would be be doubled. doubled .

---------

2

-‐ 2 ,, and and 14 11 is a constant constant ISa ec2 sec

2y+32 2y + 3z 3 2 y+ +z 2y y ++22z2 6y+3z 6y + 3z

. .

known known as as the the coefficnent coefficient of

((xx ++ 1 ) ( x- ‐ 2) 2 ) == 77xx -‐ 118 8 l )(x

friction. friction.

what is If x is the the solution solution to the the equation equation above, above, what the value the 18? value of 7x 7x -‐ 18 ?

Which of the following expresses fl in terms of the other variables? A)

/I = A) y:

a(m] a(m1 + mg) m2) "11ng2 m1m2g2

_ a(m] mz) a( m1 + + m2) B)) /In=‐ 8 "128 lg m2g -‐ m m1g

,,-

C) 14Z 111 ng2g -‐¢:7(l:t; +1112) a( m1 + 111 2) C)

-J

If the the masses masses of If were of both both object 1 and and object object 2 were doubled, how how would the acceleration doubled, would the wo acceleration of the the ttwo objects be be affected? objects affected?

informa tion . information.

1

~......._ ___

Zfi 2./x :=xx ‐- 33

m1g

Which represents all the Which of the the following following represents the possible of xx that satisfy the possible values values of that satisfy the equation equation above? above?

a(m1 + m2) - m2R

D 1)-z 4 1 1+122; ‐g ‐0m'“ g D) /1 111,

A) 1 and and 9 B) 11 and and 4 C) 4

D) 9

69

CHAPTER 8

MANIPULATING

& SOLVING EQUATIONS

------------= 2

4

x - 6x + 9

V ~

---------

Questions 31-32 refer Questions refer to the the following following 9

information. information. V v z= pP(l-r) a‐

Based on on the the equation equation above, Based the above, which which of the following could following could be be the value of x -‐ 33 ?? the value

W1

The value The V of a value V a car car depreciates depreciates over overt t years years according to the the formula according formula above, above, where where P is the the original price original price and and r is is the the annual annual rate rate of of depreciation. depreciation.

2 A); A) 3 3 B); 8) 2 7 C); C) 3 9 D); D) 2

Which of the following expresses r in terms of V,P, and t?

_I~ V A) r __ mr = 1 l - “F B) r _ = 1+

I Vp B)r‐1+\/‐P ~

Jx - = 10 = f,/x ‐- ✓-fl i m

V C) r = ~ C)r‐V‐P - 1 1

1n the equation equation above, the value In the above, what what is the value of

_P

Jx - 10? VTTE?

_

l/‘r

w D D) )r =r lz--l ‐pT

/6 A) x/E 8) ) h 2V2 B @

c >3./2 a fi C) D) v'14 \/1‐4

xy2+ X - y2-

If a car depreciates to a value equal to half its original price after 5 years, then which of the following is closest to the car's annual rate of depreciation?

1=0

A)) 0.13 A o m

lf the th e equation equation above above is true If values of true for all all real real values what must must the the value value of x be? y, what be?

B o w 8) ) 0.15

C) ) 0.16 C om D D)) o0.22

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More More Equation Equation Solving Solving Strategies Strategies 1n this chapter, In this chapter, we'll we’ll touch t w o equation equation solving solving strategies strategies that are necessary necessary for certain certain types questions touch on two that are types of questions involving equations. involving equations.

1. Matching Matching coefficients 1. coefficients 2 + 8x EXAMPLE 1: If If (x + + a) a)22 = z xx2 what is the the value EXAMPLE Bx + b, what value of b b??

It’s to see anything meaningful meaningful right right away away on on both both sides sides of the equation. Solet's let’s expand expand the the left side side first It's hard hard to see anything of the equation. So and that takes takes us and see ifif that us anywhere. anywhere.

,

( x+ + aa))2 = : (x (x+ ) ( . \ '+ + nn)) = : x 22+ 2ax+n2 (x +aa)(x + 2ax + a2

So now n o w we we have have So 7

7

’?

+ 22ax n . t+ + a2“ = z xx‘2 ++8Bx x+ x. r2‘ + + bb

both sides sides to be be equal each other, other, the the coefficients coefficients of each u s t be them up. up. For both equal to each each term term m must be equal. equal. Let’s Let's match match them

x 3 + g x + fi z x2+§x+g SO, So,

2a = B a2 = b

Solving the the equations, equations, a11z and b = : ~E. Solving = 4 and

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CHAPTER 9 MORE MORE EQUATION CHAPTER EQUATIONSOLVING SOLVING STRATEGIES STRATEGIES

Another way Another this "ma ”matching coefficients" way that that the th e SAT tests this tchin g coefficient s" strategy strategy is is to to phrase phrase the th e question question in in the the context con text of infinitely infinitely many solutions equation (we' (we’ll solutions for aa single equation ll talk talk aboutinfinitely about infinitely many many solutions solution s for for a a system of of equations next chapter). single equation the equations in the the next chapter). A single equation has infinitely infinitely many many solutions solutions only only when when both both sides sides of of the equation are equivalent. equivalent. For equation instance, For instance, 3x ++ 66 = : 3x 3x 3x + +66 has infinitely has infinitely many many solutions solutions because because no no matter matter what what the = 11 is the value value of of x x is, the the equation equation is is always always true true (x (x = is a solution a solution,, x = 2 is aa solution, z 3 is a solution, ... . . . ) Notice Notice that down to O = O, which, solution, x = a solution, that the equation eq uation boils boils down to O = 0, which, again, is always always true. true. again, Equa tion s like 3x Equations 3x + because what what’s them in in the the first place? + 6 = 3x + 6 are are a a bit bit weird weird because 's the point point of of dealing dealing with with them first place? course 3x + Of course equal to 3x + + 6! But keep keep in in mind mind that that these o t meant meant to be solved; + 6 is equal these equations equa tions are are nnot to be solved; they’re they're meant to demonstrate the concept meant demonstrate the concept of infinitely many solutions. solutions. Let’s infinitely many Let's take take a a look at at how how this this concept concept might might appear appear in an an SAT question. question. EXAMPLE 2: EXAMPLE2:

a(x2 - 2b) = 4x2 - 12 a(x2-2b) =4x2‐12

In the equation equation above, above, a and and b are constants. copstants. If the equation equation has infinitely infinitely many what is the many solutions, solutions, what value b? value of b? Just like le 1, we expand the left left side the coefficients coefficients so Just like in Examp Example expand the side and and match match the so that sides are equivalent. that both both sides equivalen t. Only equiva lent does infinitely many Only when when both both sides sides are are equivalent does the the equation equation have have infinitely many solutions. solutions.

a(x 2 -‐ 2b) = 4x 2 ‐- 12 a(x2 : 4x2 12 ax ax22 -‐ 2ab 2ab = : 4x 4x22 -‐ 12 Comparing the the coefficients, coefficients, a = 4 and 2ab = -‐12. 12. Now Comparing and -‐2ab Now we we can can solve solve for b.

-‐2ab 2ab == ‐‐12 12 2 -‐2(4)b 2(4) b = -‐ 112 b = - 12 = (TI] b=“_‐1§:‑ -8

EXAMPLE EXAMPLE 3:

kx+3(5-2x) kx+3(5 - 2x) ==15 15

In the equation above, above, I: k is a constant. constant. If the the equation true for for all all values values of x, what the equation equation is true what is the value of k ? the value ? This question is just testing you on the This question just another another way way of testing you on the infinitely infinitely many many solutions so lutions concept. concept.

kkxx + + 3(5 3 ( 5- ‐ 2x 2 x) )= : 115 5 kkxx+ +1 155-‐ 66xx = =1155 the right right side out the the x terms Since the side is just a constant constant of 15, we need need to cancel cancel out both terms on the the left left side side in order order for both sides be equivalent. easy to see that kk = does the the job. If If it helps, sides to be equivalent. It's It’s easy see that z ~ E] does side helps, you you can can think think of the the right right side as having a Ox term . The end result is th at no matter what what the value value of x is, is, 15 equa ls 15. Yes, I know as having a 0x term. The end result that no matter 15 equals know these these equations are are weird, weird, but but that’s that's how yo u get get infinitely equations how you many solutions. infinitely many soluti ons.

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Now the opposite opposite of infinitely N o w the infinitely many solutionss is is no an equation hass no solutions, are many solution no solutions. solutions. When When an equatio n ha no solution s, there ther e are no values that satisfy satisfy it. To illustrate, no values of x that illustrate, the eq equation uation

3 x++ 6 ==33xx ++110 0 3x has no solution s because has no solutions because there there is no value that can can ever make 3x equal to to 3x 3x + 10. The The equation equation itself no va lue of x that ever make 3x + 6 equal itself is aa contradiction. contradiction. This is even even more more obvious obvious if we we subtrac subtractt 3x from both sides:: we’re with 6 z 10, which both sides we're left with 6= which is is fundamentally false. fundamentally N o w in the equation + 6 = 3x + + 10, notice notice that that the x terms side have the same same coefficient coefficient of of 3, 3, but but Now equation 33: 3x + terms on each side have the the constants cons tant s of 6 and and 10 are are different. the equation to have no solutions, the coefficients of the x terms different. For an equation have solutions, coefficients the x must be must be the same same on both both sides, sides, but b u t the constants constants must must be be different. different. EXAMPLE EXAMPLE 4:

3cx Box -‐ 4(x + +1) = 2(x ‐- 1) 1) = 1)

equation above The equation above has has no no solutions, solutions, and and c is a constant. constant. What is the value value of c ‘? ?

Expanding each each side, Expanding side,

3cx 3 c x- ‐ 4(x 4 ( x++ l1)) = 2(x 2 ( x-‐ 1) l) 3cx 3 c x- ‐4x 4 x- - 44 = 22 2xx-‐ 2

= 6x ‐- 2 3cx -‐ 4 = The constants need to get the the coefficien coefficients to match. Very simply, 3c =: 66 and and cc = : [f]. I. constan ts are are different, different, so so we just need ts of of xx to match. Very simp ly, 3c

2. Clearing Clearing denominators denominators 1 1 When you solve solve an + 5xx = : 10, aa likely likely first first step step is to get r i d of of the the fractions, fractions, which are harder When you an equation equation like 5xx + is to get rid which are harder

2

3

to work work with. with. How How do do we we do By multiplying multiplying both both sides sides by times do that? that? By by 6. 6. But But where where did did that that 6 6 come come from? from? 2 2 tim es 3. So Sothis is what you're actually doing when you multiply both sides by 6: this what you're actually doing when you multiply both sides by 6: 1

1

1

1

x · (2 · 3) + x · (2 · 3) = 10 • (2 . 3) 2éx-(2-3)+%x-(2~3)=lO-(2-3) 3

£x-(Z-3)+%x~(2-Z)le~(2-3) 1.x · ('/.· 3) + jx · (2 •,3)= 10 • (2 • 3) 3 x+ +2 z 6600 3x 2xx =

We i d of the fractions fractions by clearing clearing the the denominators. denominators. Here’s the takeaway: takeaway: we same thing We got got rrid Here's the we can can do do the the same thing even even when are va variables denominators.. when there there are riables in the denominators

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CHAPTER 9 MORE CHAPTER MORE EQUATION SOLVING STRATEGIES STRATEGIES EQUATION SOLVING

EXAMPLES: EXAMPLES:

‐3 + ‐ ‐5 - == 2 -+-2 xX xx+2 +2

H a solution solution to the If x is is a 0, what what is the value the equation equation above above and and ix > 0, is the value of of x ? ? same way way we In the the same wemultiplied before,, we we can multiply multiplied by by 2 ·~3 before can multipl y by by x(x x(x + 2) 2) here. here.

3 xg-x(x+2)+x+2-x(x+2)=2-x(x+2) ·x(x + 2) + x +5 2 · x(x + 2) = 2 . x(x + 2)

3 5 g-x(x+2)+xj(z-le/+47=2X(X+2) .j(x + 2) + .x-rt •x..(x.-+-2f= 2x(x + 2) 1

3(x+2)+5x=2x2+4x 3(x + 2) + Sx = 2x 2 + 4x

3x+6+5x=2x2+4x 3x + 6 + Sx = 2x 2 + 4x 0 = 2x 2 - 4x - 6 0:2x2‐4x‐6 0 = x 2 - 2x - 3 Ozxz‐Zx‐B

0 :=( x(x‐ -3 3)(x ) ( x ++1l )

=.. GJ.

x= l but because x > 0, z B3 oorr x = -‐1butbecausex O,x x=

Here 's one one final example example that that showcases showcases both both of the the strategies Here’s strategies in this thi s chapter. chapter.

EXAMPLE 6: EXAMPLEG:

3x + 55 _ -‐6x2+11x+5 3x 6x2 + llx + 5

axx+ (x+l)(ax+2) xx+1 +1+ a +2 = _ (x+1)(ax+2)

equation above, above, 1:75‐g x f:. - ~ and the value In the the equation and a is aa constant. constant. What What is the value of a ?7

a

A) ‐6 -6

B) -‐22

C) 2

D) 6

Let's clear clear the the denominators denominators by multiplying multipl ying both Let's both sides sides by (x + + l1)(ax )(ax + + 2):

3x 55 3x -‐6x2+11x+5 6x2 + llx + 5 x+1~~~(x+1)(ax+2)+ax+2(x+1)(ax+2)‐mm(x+l)(ax+2) --·(x+ l )(ax + 2)+-- (x+ l )(ax + 2)= ( l )( ) •(x + l )(ax + 2) x +l ax + 2 x + ax + 2 3x 5 3x 5 -‐6x2+11x+5 6x + ll x + 5 m -M ( a x ax + +22)) ++ ~ + 1 =) ~ M _ m . ·~ x ~ •.(,x-l-'11( •-(x(+ x l ).{ax-+-2T 2

3x(ax 3 x ( a x++ 22)) + 5S(x ( x ++ 1 1)) := -‐6x2 6x 2 + 11 llx x++ 5 2 3ax2+6x+5x+5z 3ax + 6x + Sx + 5 = ‐6x2+11x+5 - 6x 2 + 1lx + 5

[ill].

2 terms, Comparing the the coefficients coefficients of the xx2 Comparing terms, 3a = -‐ 66.. Therefore, Therefore, a = = ‐- 22.. Answer Answer (B) .

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CHAPTER EXERCISE:Answers for this chapter start on page 293.

A calculator should NOT be used on the following questions.

1

2

18x2 - 8

3

2

3

b)

the equation equation above, above, a and are constants. constants . In the and b are ? Which the value Which of the following following could could be be the value of ab ab?

6x2 + ~x) 3x) = : M3 bx2 + ex cx (x ~x ax + bx

30(x3 + 30

= 2(ax + b)(ax -

2

A) 6 A) B) 9

In the re constants. the equation equation above, above, a, b, and and ccaare constants. If the equation true for aJI the equation is is true all values values of x, what what is the va lue of a + b + cc?.7 thevalueofa

C)) 112 C 2 D)) 3366 D

xx - ~2(33c-+‐8)‐22 4x (3x + 8) = 2 ( 2 - ~x)

1

-(‐1)

How many many solutions are there there to the equation How solutions are the equation above? above? The equation equation has no solutions. solutions. A) The has no

2

B) The The equation equation has infini tely many many so lution s. B) has infinitely solutions. C) The The equation equation has has exactly exactly 11solution. solution. C) D) The has exactly solutions. The equation equation has exactly 2 solutions.

ln the equation above, a and and b bare ts. If In the equation above, are constan constants. If the equation has solutions, the equation ha s infinitely infinitely many many so lutions, what what

is the value ? the va lue of ~g-? 3

A) 4

3 ( 3-‐ 22x) x ) = 1122 -‐ 77xx 3xx + aa(3

4

B) -

If the In the the equation equation above, above, a is a constant. constant. If the equation solutions, what equation has has no no solutions, what is the the value value of aa??

3

C) 6 D ) 112 2 D)

A A)) -‐ 2 B) 2

C C)) 4 D) 5

ax - b = 3(2x + l ) In the the equation equation above, above, a and b are constants constants.. If ln and bare the equation has no solution, which the the equation has no solution, which of the following be the values of aa and and bb ?? following could could be the values

If (2): 5) = = 12x 12x22 + bx bx -‐ 15 (2x + 3)(ax -‐ 5) 15 for all all values value values of x, what what is the the va lue of b ??

A ) aa = z 22 aand n dbb= = ~- 3 A)

B =2a n dbb= = 33 B)) aa= and C a n db b=: -‐ 3 C)) aa=: 6 and

A) 6

D ) aa == 6 a n dbb== 3 D) and

B) 8 C 0 C)) 110

D 2 D)) 112

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CHAPTER 9

MORE EQUATION SOLVING STRATEGIES

If (x + 3y) 2 = x 2 + 9y2 + 42, what is the value of x2y2?

IIff n < Oand + 9 = (2x (2x+n)2, what is 0 and 4x24» 4x 2 + m mxx+ + n)2, what is the value o thevalue + n ?? off m + A) - 15

8)) ‐-‐ 9 B C)) -‐ 3 C D)) 112 D 2

1 If ~ + ~ = ~, what is x in terms of p and y ? If‐+1= l,whatisxintermsofpandy? p x yy P

6 x = xx ‐-33x(2n x ( 2 n-‐ 11)) 6x

X

In the the equation constant. If equation above, above, n is a constant. If the the equation has equation solutions, what what is has infinitely infinitely many many solutions, the value value of n ?? the

A) 19‐3! p- y A) 3) fly 8) _EL

p+y

A) -A) ‐523 8) -B) ‐531 C) C)

(C) D PY j l p- y

D) _EL W D) y- p

4

3 5

D) D) -

3

(x3+kx2‐3)(x‐2)=x4+7x3‐18x2‐3x+6 (x 3 + kx 2 - 3) (x - 2) = x 4 + 7x 3 - 18x 2 - 3x + 6 In the constant. If It the the the equation equation above, above, k is a constant. equation values of x, what is the equation is true true for all values what is the value of k? k? value

in

ab + a

a

.

If = g + 5 for all values of b, what If -ab2b- a = b+ all values what is 1s the the

A A)) -‐ 99

value value of a a??

B) 8) 5 C) 7

D D)) 9

1 If 1,what thevalue x? If~i -‐ ‐y4‐ - - = 1, what is the value o off x? X

X - 4

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THE THE COLLEGE PANDA COLLEGE PANDA

5

ax - b xx ++33 ' xx -- 22 ‘ (x ( x + 33)(x ) ( x-‐ 22)) 2

The equation equation above The above is true true for for all all x > 2, 2, where where a and b bare constants. What and are constants. What is the the value value of a + + b ??

A) 7 B) 13 B) 13

C)) 119 C 9 D)) 221 D 1

4

+

2

_

35 35 - 1

--+--=-x -‐ 1l xX + 1 1 _x2‐1 x2

If x > the solution > 1, 1, what what is the the equation solution to the equation above? above?

The equation (2x - b) (7x + b) = 14x 2 - ex - 16 is tme for all values of x, where band care constants. If b > 0, what is the value of c ? A) - 20 8) 20

C) 28 D) 36

+ --3!!.__ =3 n‐l+n+1= n- 1 11+ 1

_ 3_

If n > 0, for what If what value value of of n 11is is the the equation equation above true? above true?

77

Systems of Equ Systems Equations ations A system refers to more equations equations that that deal deal with with the same set set of of variables. variables. system of equations equations refers to 22 or or more the same

+ yy =z ‐77 -‐5.t sx + 2}; = z -‐ 112 2 -‐ 33xx -‐ 2y There are w o main main ways ways of solving solving systems of 22 equations: equations: substitution There are ttwo systems of substitution and and elimination. elimination.

Substitution Substitution Substitution is all Substitution all about isolating one variable, or y, y, in in the way possible possible. about isolating one var iabl e, either either xx or the fastest fastest way

Taking the example we can can see see that it’s easiest easiest to to isolate in the equation because because it it has has no Taking the example above, above, we that it's isolate y y in the first first equation no coefficient. Adding 5x to both both sides, sides, we we get get coefficient. Adding Sx : Sx 5x -‐ 7 y= We can can then then substitute substitute they the y in second equation equation with with Sx 5x -‐ 7 We in the the second 7 and and solve solve from from there. there. 7 )=: -~12 -‐ 33xx -‐ 2(5x 2(5x -‐ 7) 12 10x + + 14 1 4z -‐ 33xx -‐ lOx = ‐- 1122 3x = : -‐ 226 6 -‐ 113x z 2 x=

Substituting xx =z 22 back intoy : Sx 5x -‐ 7,y : 5(2) 5(2) -7 77 = z 3. 3. Substituting back into y= 7, y = The z 2,}/ 3, which can be be denoted denoted as as (2,3). (2, 3). The solution solution is x = 2,y z= 3, which can

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Elimination Elimination Elimination is is about about getting getting the the same same coefficients coefficients for for one one variable t w o equations you can can add Elimination variable across across the the two equations so so that that you add

or subtract the the equations, equations, thereby thereby eliminating that variable. variable. or subtract eliminating that Using the the same same example, example, we we can can multiply multiply the the first first equation equation by so that that the y’s have have the coefficient (we Using by 2 2 so the y's the same same coefficient (we don’t worry worry about about the the sign sign because because we we can can add add or subtract subtract the equations). don't equations). 0x + +Z = -‐ 114 4 -‐ 110x 2yy =

-‐ 33xx -‐ 2y 2 y:= -‐ 112 2

eliminate y, we we add add the equations. To eliminate the equations. 2y = = ‐- 114 4 -‐10x 10x + 2y -‐ 33xx -‐ 2 2yy = : -‐ 112 2 2 -‐ 226 6 -‐13x 13x = Now, we we can can see see that that x = = 2. This This result result can can be be used used in either either of the original equations to solve solve for y. We’ll Now, original equations We'll pick pick the the first equation. equation.

+ 22y y= : -‐ 114 4 -‐10(2) 10(2) +

2 y=-: ‐ 114 4 -‐ 2200++ 2y 2 =6 2yy = 6

yy = = 33 And we got substitution: x = 2, y z = 3. And finally, we we get get the the same same solution solution as aswe got using using substitution: : 2,y 3. When solving solving systems systems of equations, equations, you you can can use use either either method, one of them will typically be faster. If If When method, but but one them will typically be you see a variable variable with w i t h no no coefficient, coefficient, like like in -‐ 55xx + + y = z -‐77 above, substitution is likely likely the the best you above, substitution best route. route . If you see matching matching coefficients coefficients or you you see that that it's it’s easy easy to get matching coefficients, likely the best you get matching coefficients, elimination elimination is likely the best route.. The The example example above above was was simple simple enough enough for both both methods methods to work work well (though substitution was slightly route well (though substitution was slightly personal preference. preference. faster). faster). In these these cases, cases, it comes comes down down to your your personal

No solutions solutions In the solutions when when both the equation equation are the previous previous chapter, chapter, we we saw saw that that aa single single equation equation has has no no solutions both sides sides of the are the the same except except for the the constants. constants. same

similar fashion, fashion, aa system system of equations equations has has no no solutions solutions when the ttwo w o equations equations are are the the same except for their their In similar when the same except constants. For For example, example, the the system system constants.

3 x+ + 2y 2 y=z 5 3x 3 x++ 2y 2 y== -‐ 4 3x has no solutions solutions since since the the different different constants constants (5 (5 vs. vs -‐ 44)) result result in equations contradict each each other. has no equations that that contradict other . There There isn’t an an x and and aa y that that can can possibly possibly satisfy satisfy both both equations equations at the same time.. Note Note that the system system same time that the isn't

3 x+ +2 = 55 3x 2yy = 6 x++4y 4 y=: -‐ 8 6x also has has no no solution. solution. Why? Because the second equation equation can divided by 2 to get the the contradictory contradictory equation also Why? Because the second can be be divided equation we had had before. before. we

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CHAPTER EQUATIONS CHAPTER 10 10 SYSTEMS SYSTEMS OF OFEQUATIONS

EXAMPLE EXAMPLE 1: ‐ 1 i j= = 15 15 --‐ax ax-12y 41: + By = = -‐22 4x+3y

If the system system 0f of equations equations above above has has no no solution, solution, what what is the the value value of a ? 11 ? We must get get the the coefficients coefficients to match match so so that that we we can can compare compare the w o equations. do that, we multiply the We must the ttwo equations. To do that, we multiply the second second equation equation by by -‐ 44::

-‐ aaxx -‐ 12y 123; = z 15 -‐16x 16x -‐ 12y 1211 = = 8 See how the -12's match now? then we we get get o our contradicting equations equations how the ‐12’s match now? Now N o w let's let's compare. compare. If aa = ~ then u r two t w o contradicting with with no no solution. solution. One One constant constant is 15 15 while while the the other other is 8. 8.

Infinite Infinite solutions solutions In previous chapter, chapter, we we learned infinitely many when both both sides In the the previous learned that that aa single single equation equation has has infinitely many solutions solutionswhen sides of the the equation equation are are the the same. same.

Similarly, both equations equations are are essentially essentially the the same: Similarly, aa system system of equations equations has has infinitely infinitely many many solutions solutions when when both same:

3 x+ + 2y 2 y=z 5 3x 3 x+ + 2y 2 y= = 55 3x (1, name just a few. Note that the the system (1, 1), 1), (3, (3, -‐ 22), ) , (5, (5, -‐ 55)) are are all all solutions solutions to to the the system system above, above, to name just a Note that system

6 x++ 4y 4 y=: 110 0 6x 3 x++2y 2 y=: 5 3x 5 also divided by by 2 to to get get the They're also has has infinitely infinitely many many solutions. solutions. The The first equation equation can can be be divided the second second equation. equation. They’re still still essentially essentially the the same same equation. equation.

EXAMPLE 2: EXAMPLE2:

Bx -‐ 5y 531 = = 8 3x 8 mx -‐ ny = 32 mx 32

In the system equations above, If the the system system has many solutions, what In,the system of equations above, m and and n are are constants. constants. If has infinitely infinitely many solutions, what isthevalueofm+n is the value of m + n?? Both the same there to be many solutions. solutions. We multiply the the first equation equation by Both equations equations need need to be be the same for there be infinitely infinitely many We multiply 4 to get hand sides get the the right right hand sides to match match::

12x -‐ 20y = 32 12x mx -‐ ny 71}; mx

=

= 32 32

N o w we can can clearly clearly see see that that m m = 12 12 and and n = : 20. Therefore Therefore,, m + + n= : -. Now I32 I.

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Word Word problems problems You will most u n into into a question asks you into aa system system of most definitely definitely rrun question that that asks you to to translate translate a a situation situation into of equations. equations. Here's Here's a a classic example: example:

EXAMPLE order lunch lunch from aa restaurant. restaurant. Each EXAMPLE 3: A group group of 30 students students order Each student student gets gets either either a a burger burger or or asalad. thepriceofasaladis$6. Ifthegroupspentatotalof$162,how a salad. Thepriceofaburgeris$5and The price of a burger is $5 and the price of a salad is $6. If the group spent a total of $162, how many many students students ordered ordered burgers? burgers? Let x be the number burgers and can then then be the number of students students who who ordered ordered burgers and yy be be the the number number who who ordered ordered salads. salads . We We can make w o equations: make ttwo equations:

0 x + y z=330 5 62 5xx +66yy ==1162 Make Make sure sure you you completely completely understand understand how how these these equations equations were were made. made. This This type type of of question question is is guaranteed guaranteed to be be on the the test. test. We'll use We’ll system. Multiply Multiply the the first and subtract: use elimination elimination to solve solve this system. first equation equation by by 6 6 and subtract:

6x+6y 6x + 6y == 180 5x+6y : 162 5x + 6y =

x=‑ X=~

18 students got 18students got burgers. burgers.

More complex complex systems systems You might might encounter bit more encounter systems systems of equations equations that that are are a a bit more complicated complicated than than the the standard standard ones ones you’ve you've seen seen above. these systems, and some equation manipulation w i l l typically do the trick. above. For these systems, substitution substitution and some equation manipulation will typically do the trick .

EXAMPLE4: EXAMPLE 4: y + 3x = 0 y+3x x2 + 2y 2y22 = = 76 x2

If If (x,y) (x,y) is a solution to the y? a solution the system system of of equations equations above above and and y y > 0, 0, what what is is the the value value of of y? ln In the the first equation, we isolate isolate y to get get y = : -‐ 33x. x . Plugging Plugging this equation, we this into into the the second second equation, equation, 2 2(‐3x)2 xx2+ + 2(3x) 2

: 76 76 =

2 + 2(9x2) xx2 2(9x 2 )

= 76 76

xx2+1sz2 + 18x2

= 76 76

2

2 19x2=76 19x = 76

x2=4 x2 = 4 xX = : i± 2 IIff xx = 2, : ‐3(2) thenyy = = ‐- 33((-‐ 22)) := 6. > 0,y : ~ IE].2, theny then y = - 3(2) = = ‐- 66.. IIff xx = -‐ 22,, then 6. Becausey Because y > 0, y =

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OF EQUATIONS SYSTEMS OFEQUATIONS 10 SYSTEMS CHAPTER 10 CHAPTER

EXAMPLES: EXAMPLE 5: xy + 23; = 2 xy+2y=2

) - 6= 0 (-1 )2++(x+2) (_1 x+2 ‐ 6 _ 0 (x+2) x+2

1

2

1

IYI?? value for |y| a possible what is a above, what equation above, the equation (x, y) is a solution If (x, If solution to the possible value substitution clever substitution be a clever might be there might that there hint that This is a hint both equations. + 2)'s the (x -+Notice the Notice 2)'s lying around around in both equations. This equation, the Isolating one. this as complicated as problem a somewhere, especially for a problem ascomplicated as this one. Isolating y in the first equation, especially somewhere,

xy + 2y = 22 y(x + 2) 2) = z 22

_ 2 y _ x+2 1 1 y__ with z. equation with second equation the second substitute - - in the can substitute this form? want this would I want Why would y_ = here, 1 From here, From z -‐. - . Why form? So 50 I can 2 x+2 x+ 2 2 this as this such as manipulations such simplifying manipulations any simplifying out for any an eye must questions, you tougher questions, these tougher do these you do As you As you m u s t keep keep an eye out one. one.

Substituting, we get Substituting, we 1 ‘ + (- 11 ) - 6 = 0 )2 _1 ((m) x+2 x + 2 +(x+z)‘6=° ’)

)2+ G)- = o (f owe‐6:0 6

+ Y-.- 6 = 0 y2 g+g~6zfl 4

2

f+@‐M=O + 2y - 24 = 0 y2 ((yy+ + 66)(y ) ( y- ‐ 4) 4 )== 0 Finally,y z ‐6 |y| can either /‑6 or 4 /. be either can be and IYI 4, and or 4, - 6 or Finally, y = even then, And even Practice. And use? Practice. can use? you can "trick" you substitution or ”trick” clever substitution there's a clever know whether you know will you How H o w will whether there’s then, a without a done without be done designed to be are designed questions are that SAT questions keep in mind Just keep sure. Just know for sure. always know won't always you won’t you mind that try and back step a take waJI, a hitting or circles in running you're like feel you if So steps. of number crazy crazy number steps. 50 you like you’re running circles hitting a wall, take a step back and error. and error. trial and with trial comfortable with be comfortable you must score, you perfect score, a perfect else. To get a something something else. must be ? a possible positive what is apossible = 10, and yz = 5, and = 5, 8, xz = = 8, If xy = EXAMPLE6: If EXAMPLE 10, what positive value value of xyz ?

result right sides multiply the right and multiply sides, and the left sides, Multiply the equations. Multiply three equations. Multiply all three trick . Multiply Here 's the trick. Here’s sides.. The The result is 2y2z2 = = 8 -· 5 ~10 - 10 xchyzz2

xzyzzz = 400 x 2y2z2 = both sides. Square o o t both sides. Square rroot

= ± v'400 x2y2z2 /W = HEW) = i±220 0 xyz = answer Since the value, the Notice how get the answer able to get were able we were how we / 20 /. Notice answer is -. the answer positive value, a positive asks for a question asks the question z. x, y, or 2. values of x,y, without knowing individual values the individual knowing the without 82

THE COLLEGE PANDA THE COLLEGE PANDA

Graphs Graphs Leaming our understanding understanding of systems Learning aa bit bit about about equations equations and and their their graphs graphs will will inform inform our systems of equations equations.. The solutions solutions to a system system of equations equations are the intersection intersection points points of the graphs equations. Therefore, The graphs of the the equations. Therefore , the number points . number of solutions solutions to a system system of equations equations is equal equal to the number number of intersection intersection points.

Take, for example, example, the system system of equations equations at at the the beginning beginning of this chapter: chapter:

-‐5x 5x + yy == -‐77 -‐ 33xx -‐ 2}; 2y = -‐ 112 2

We can put put both both equations equations into into y = mx mx + ‐+- b bform form (we (we won't won’t show show that that here) here) and get the the following We can and graph graph them them to get following lines lines.. y

(23)

The one intersection point, so so there The solution solution to the system, system, (2,3), is the the intersection intersection point. point. There There is only one intersection point, there is only only one solution one solution.. What What about about graphs graphs of systems systems that that have have infinite infinite solutions solutions or no solutions? solutions? Graphing the the following equations contradict contradict each Graphing following system, system, which which has has no no solution solution because because its equations each other, other,

y - 22xx=: 1 l z ‐- 3 y -‐ 22xx = weget we get y}/

What points . They're parallel. Makes What do you you notice notice about about the the lines? lines? They have have no intersection intersection points. They’re parallel. Makes sense, sense, right? right?

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CHAPTER EQUATIONS CHAPTER 10 10 SYSTEMS SYSTEMS OF OFEQUATIONS

And with infinite infinite solutions? And for a system system with solutions?

2 y-‐ 4x 4 x= z 22 2y y -‐ 22xx = 1 y

It's just one it’s two t w o lines, lines, but but because because they’re overlap and It's just one line! Well, actually actually it's they're the the same same line, line, they they overlap and intersect intersect in in an Hence, an an infinite infinite number an infinite infinite number number of places. places. Hence, number of solutions. solution s.

EXAMPLE 7: In the the xy‐plane, the lines lines y = 3x 3x -‐ 5 2x+ at the the point point (h,k). (h,k). What What is is EXAMPLE 7: xy-plane, the 5 and and y = = ‐-2x +1100 intersect intersect at the value ofk? of k ? thevalue As mentioned mentioned earlier, solutions to aa sys system equations are the graphs those ear lier, the the solutions tem of equations are the the intersection intersection points points of the grap hs of those eq uations, and equations, and vice vice versa. versa. So So to find the point(s) point(s) where where two graphs intersect, solve solve the system system consisting graphs intersect, consisting of this problem, their equations. equations. 1n In this problem, that that system system is

: 3x y= 3x -‐ 5 : ‐2x y =2x++ 10 10 Substituting the the first equation Substituting equation into we get into the the second, second, we get

3 x-‐ 55 = x ++110 0 3x =‐-22x 5 x = 1 5 Sx = 15

x =3 When x =z 3, y = 3(3) -‐ 5 = 4. So So the w o lines lines intersect intersect at (3,4) When the ttwo (3,4 ) and and k =z I. [i].

84

PANDA COLLEGE PANDA THE COLLEGE THE

EXAMPLES: EXAMPLES:

y=xz‐5x+6 y = x 2 - 5x +6 yy=x+1 =x+1

intersection an intersection represents an y) represents (x, y) pair (x, ordered pair the ordered If the xy-plane. If the xy-plane. graphed in the above is graphed equations above system of equations The system The ? value of y ? possible value one possible what is one equations, what the ttwo graphs of the point point of the the graphs w o equations, equation Substituting the first equation system. Substituting solve the system. let's solve so let’s points, so intersection points, the intersection are the the system solutions to the The solutions The system are second, we get into the second, into

xz‐Sx+6:x+1 x2 - 5x + 6 = x + 1 2 - 6x + 5 = 0 xx2‐6x+5:0 ( x-‐ 11)(x-5)= )(x-5)=0 (x x ==110orr 55 (1, 2) intersect at (1,2) equations intersect the ttwo graphs of the So the graphs When x = When = 1, 1,yy = l1 + +11 = : 2. When When x = : 5, 5,yy =: 5 + +11 = = 6. So w o equations

[I]. and E. possible values which means (5,6 ), which and (5,6), and means the possible values of y are[}] are and EXAMPLE9: EXAMPLE9:

y2=x+3 y2 = x+3 y ==| lxl xl yy

does solutions does many solutions How above. H shown above. xy-plane are shown the xy-plane graphs in the and their equations and system of ttwo A system w o equations their graphs o w many the system system have? have? A) One A)One

B)Two B) Two

Three C) Three

D)Four D) Four

Answer ~-(B) . solution s. Answer are ttwo there are so there places so intersect in ttwo graphs intersect Simple. w o places w o solutions. Simple. The graphs

85 85

CHAPTE R 10 CHAPTER SYSTEMS OFEQUATIONS 10 SYSTEMS OF EQUATIONS

CHAPTEREXERCISE:Answers for this chapter star t on page 296.

_

A calculator should O T be the should N NOT be used used on the following questions. following questions .

2 + 5Syy = 2xx+ =2244 x+4y= x + 4y = 15 15

3 x-‐ 5 = ‐- 1111 3x Syy = lX e= ‐1 -B 3yy

If above,, If (x, (x, y) y ) satisfies satisfies the the system system of equations equations above what what is the the value value of x + + y ?? A)) 7 A

What What is the the solution so lu tion (x, (x,yy)) to to the the system sys tem of equations above? equations above?

B) 8

C) 9 D)) 110 D 0

A)) ((-‐ s5,2 A ,2) B)) ((-‐ z2,1 B , 1) C) (1,0) (1,0) C) D) ((4, - 1) 4,‐1) D)

_ _

3 x++yy = = ‐- 22xx + 8 3x 0 -‐ 33xx + 2 2yy = --110

y+ : 2200 + 22xx = 6 y ==112 2 6xx-‐ 5Sy

If (x, y) system of equations equations 1f y) is a a solution sol u tion to the system above, w hat is the above, what value the va lue of xy ?? A)) ‐- 116 A 6

Wh at is the the solution What solution (x, system (x,yy)) to the the sys tem of equations above? equations above?

B)) -‐ 8 B

A ) ((-‐ 77,6 ,6) A)

C)) -‐ 4 C

B) (‐6, B) (- 6, 7)

D)) 44 D

C) C) (6,7) (6, 7) D) D) (7,6) (7,6 )

x+ y z=aax + bb

y z= ‐-bbxx

3x -‐ 4y = = 21 2] 3x

The w o lines xy-plane The equations equations of ttwo lines in the the xy-p lane are are shown above, where a and b are constants. If the the shown above, where and bare cons tants. If ttwo w o lines value lines intersect intersec t at at (2,8), (2, 8), what what is the the va lue of aa?7

=

4x 14 4x -‐ 331 3y = 14

IfIf (x, (x, y) is a a solution solution to the the system system of equations equations above, above , what w hat is the the value value of y -‐ x ? ?

A A)) 2

A 8 A)) ‐- 118

B) 4

B B)) -‐ 5

C C)) 6

C) 5

D) 8

D) 8

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THE COLLEGE COLLEGE PANDA THE PANDA

yy=x2+1 = x +1 z xx -‐ 1l y= 2

2xx-‐ 4 4yy= 8 2 z8 xX + + 22yy = 4 How many many solutions are there the How solutions (x,y (x,y)) are there to the system of equations above? system equations above?

y

A) Zero Zero A)

One B) One C)) Tw Twoo C

More than D) More than ttwo wo

2x -‐ 5}; Sy = a bx + lOy 10y = ‐8 - 8

A system equations and their graphs system of ttwo w o equations and their graphs in

the man y the xy-plane xy-plane are are shown shown above above.. How H o w many solutions solutions does does the the system system have? have?

ln the the system system of equations equations above, above, a and and b bare In are constants . If the system has infinitely many constants. If the system has infinitely many solutions, what what is the the value solutions, value of a ??

A) Zero Zero A)

One B) One C) Twoo C ) Tw

A ) -‐ 4 A) 4

D) Three Three

1 B) 3 4

C) 4 D 6 D)) 116

= y +2 2 -‐ 5Sxx = 2 ( 2 x- ‐1)1 )=: 3 -‐ 33yy 2{2x What What is the the solution solution (x, y) to the the system system of equations above? equations above?

a x + 2y 2 y=z 5 ax+ 5 3 x-‐ 66yy ==2200 3x

A ) ({-‐ 22,8 ,8) A)

the system above, a a is a a constant. constant. In the system of equations equations above, the system system has one solution, solution, which the If the has one which of the following can can NOT be the the value value of aa ?? following N O T be

B ) ( -‐ 11,, 33)) B) C ) ((1, 1 ,-‐ 77)) C) (3, - 17) D) (3,‐17)

A A)) -‐ 1 3 B) 3 B) -

E 4

C) 1 D D) ) 3

87

CHAPTER 10 EQUATIONS CHAPTER 10 SYSTEMS SYSTEMS OF OFEQUATIONS

3x = 15 3x -‐ 63) 6y = 15

= -8 4x‐‐y=‐8 3

1 4x - - y

‐-2x 2x + 4 4yy= = -‐ 1100

x ++116 6 y ==44x

How solutions (x,y) are there H o w many many solutions (x,y) are there to the the system of equations equations above? above? system

What the solution the system What is is the solution (x,y) (x,y) to to the system of equations above? equations above?

A) Zero Zero A)

A) A ) ((-‐ 22,8) 8) 8))( (- 1,12 1,12)) C)) , (1,20) ( 120) D)) ( (3,28) 3 , 28)

8) One B) One C)) Tw Twoo C

More than than ttwo D) More wo

mxx-‐ 66yy = = 1100 m 2xx-‐ n nyy = 2 z5

0 .0.5x 5 x ++114 4 y := x -‐ yy ==‐ -118 8

_

X

ln the system system of equations equations above, above, m m and and n are are In the the system system has constants. If constants. If the has infinitely infinitely many. many

According above, According to the the system system of equations equations above, what is the value of y ? what the value ?

.

.

m

so lutions, what what is the value m ?? solutions, 15the value of g 11

1 A) 12 fi A)

1 B)§ 8) 3

4 C) C) 5 3 D D) ) 3

1l 1 3x y ‘ -a6y l l ‐= 44

6 x-‐ aayy = =8 6x

:\/E+3 y= vx + 3

ln equations above, [n the the system system of equations above, aa is aa constant. constant. If the the system the value system ha hass no no solution, solution, what what is the value of aa?? A)

m_ fu -

y =3

(x, y) is the the solution solution to the the system system of equations equations IfIf (x, above, the value value of y ?? above, what what is the

1 3

8) B) 1

C) 3 D) 6

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PANDA THE COLLEGE PANDA THE COLLEGE

allowed on the following calculator is allowed A calculator following questions. questions. B

0 @

medium, small, medium, sells jelly in small, supermarket sells A local local supermarket much as weigh Sixteen small and large jars. jars. Sixteen small jars jars weigh as much and large Four small large jar. Four jars and medium jars as w o medium and one one large small as ttwo weight same weight the same have the medium jar have one medium jars and one jars and the small jars many small How many large jar. How as jars have have the one large as one large jar? one large weight weight of one A) 7 A) 7 B) 8

depending on points depending A game on rewards points darts rewards game of darts and regions, A and are ttwo There are which w o regions, hit. There region is hit. which region hitting darts, hitting throws 3 darts, above . James shown above. B,as James throws B, as shown total of a total twice, for a region B twice, and region once and region A once region hits but hits darts, but throws 33 darts, 18points. also throws Oleg also points. Oleg 18 regions once for a total of a total region B once and region twice and A twice regions A rewarded for are rewarded many points How many points. How 21 points. 21 points are once? hitting region B once? hitting region

B) 8

9 C) 9 D) 10 10

are points are 30 questions, 5 points with 30questions, test with math test a math On On a points correct answer rewarded answer and and 2 points each correct rewarded for each answer . If James incorrect answer. each incorrect are James deducted for each are deducted points, scored 59 and scored questions and answered 59 points, the questions answered all the systems of following systems the following which of the solving solving which answers, correct answers, number of correct his number equations gives his equations gives the answers, incorrect of x, and incorrect answers, y, on on the number his number and his

rectangular tables, rectangular A restaurant types of tables, two types has two restaurant has circular ones people and and circular seat 4 people each seat can each that can ones that people If 144 people tables people. It seat 8 people. each seat can each that can tables that restaurant, the restaurant, at the tables at aU 30 tables are enough to fill all are enough restaurant the restaurant does the tables does how rectangular tables many rectangular how many have? have?

math test? math test?

A)

B) B)

59 xX + +y y = = 59 5x -‐ 2y Zy = = 30 30 5x x X

+ = 30 + y!f =

A) 12 A) 12

5x = 59 59 2y = 5x + 2}; c) C)

B) 16 16 B)

C) C) D) D)

x+ +y = = 30 2x Sy = 59 2x ‐- 53,

D D)) xX++yy ==3300 5 x-‐ 22yy =: 559 9 5x

89

20 20 24 24

CHAPTER 10 SYSTEMS OF EQUATIONS

In the 7x + +7 the xy-plane, xy-plane, the the graph graph of y = = x2 x 2 ‐- 7x 7 intersects the the graph intersects graph of y = : 2x the points 2x -‐ 1 at the points and (p,q). (p, q). What (1, 1) 1) and What is is the the value value of p p??

y

_

xx22- ‐2 2x x==yy-‐ 1 1 x = y ‐- 111 1

A system system of ttwo equations is graphed graphed in the the w o equations xy-plane above. Which of the following is the xy-plane above. Which the following is the solution (x,y) the system? system? solution (x, y) to the

If (x, (x, y) is a a solution solution to the the system system of equations If equations above, what what is one one possible possible value above, value of y ??

(0, - 6) A) (0,‐6) A) 8)) (B ( ‐ 33,,-‐ 33))

C) ( - ~, - 3) c> (4-3) D) (-3, -;) r» (+2)

x2 _ x2 -

1 y2=_ -11_2

yz

12 xX -‐ 2y : 00 2y =

lf the the ordered Y1) and satisfy If ordered pairs pairs (x1, (x1,y1) and (x2, yi) y2) satisfy the system system of equations equations above, above, what what are are the the the values of y1 and values and 1/2 y2 ? ? 1 and 2 1 1 B ‐ ‐ ‐ a and n d -- ‐ ‐ B) ---

A)

--

1

2

\/12 v'12

)

12 v'12

1 1 -- and and E C) C) ‐i 4 4

D) D)

--

1

6

1 and and1 6

90

ll

Inequalities Inequalities Just had equations equations and and systems systems of of equations, equations, we we can and systems systems of of inequalities inequalities.. Just aswe as we had can have have inequalities inequalities and The you m u s t reverse reverse the the sign sign every time you multiply or or divide sides by The only only difference difference is that that you must every time you either either multiply divide both both sides by aa negative negative number. number.

For For example, example,

2x + + 33 < 99 2x reverse the point? Well, we we would subtract by by 33 to to get get 2x and then by Do we we have have to reverse the sign sign at at any any point? would subtract 2x < 6 6 and then divide divide by 2 to get get x < 3. Yes, we we did did aa subtraction subtraction but but at at no no point point d i d we we multiply multiply or or divide by aa negative negative number. did divide by number. Therefore, the the sign stays the the same. same. Therefore, sign stays Let’s take Let's take another another example: example:

3x ++ 55 < 4x + 4 3x The first step step is to combine combine like like terms. terms. We Wesubtract both sides by 4x to get x’s on the left left hand We then then The subtract both sides by 4x to get the the x's on the hand side. side . We subtract subtract both both sides sides by 5 to get get the constants on the the right right hand the constants hand side: side :

3 x-‐ 4 3x 4xx -‐xx -‐ 1 represents above, the inequality region above, shaded region As shown points above above the line shown by the shaded look at below, just what's below, and what’s line and a line above a what's above track of what’s keeping track time keeping hard time a hard have a y= just look at the you have If you - x ‐- 1. If = ‐x region. The "above" region. the ”above" always in the y-axis is always part of the y‐axis two parts into two cuts the y-axis into y-axis. The line cuts parts.. The top part with intersection with show the intersection doesn't show graph doesn’t "below" region. always in the ”below" part of the y-axis is always bottom part bottom region. If the graph and "above" and the ”above” determine the graph to determine the graph through the vertical line through draw your always just draw the y-axis, you your oown w n vertical you can always regions. "below" regions. “below” itself do line itself the line points on the - x -‐ 1, the points and NOT y z= ‐x Because y > -‐xx -‐ 1 and dashed . Because line is dashed. that the line note that Also note line on points and solid, be would line the then 1, x 2:: y were inequality . If the equation the inequality. satisfy the not n o t satisfy equation were 2 ‐x ‐ then line would be solid, and points on the line inequality . the inequality. satisfy the would satisfy would y

- X

example, inequalities? For example, system of inequalities? a system about a what about But what

y ::; 5 ‐x - x ++ 44 1

-x- 3 yY > - ~2 x ‐ 3 inequalities. In this both inequalities. satisfy both that satisfy points that with the points region with goal is to find the region comes to graphing, When it comes When graphing, the goal 1 . . can we can points, we this set of pomts, locate this = ~x y= above 31 but above below y = -‐xx + that are below points that want the points we want case, we + 4 but Ex ‐- 3. 3. To locate

1x-

-x

1 . overlap . the rregions where the and see where and above = ‐x ++ 4 and below y z regions below shade the regions shade above y = Ex -‐ 3 and e g i o n s overlap. y

X

system. solutions to the system. that are solutions points that the points contains all the the left contains The overlapping on the region on overlapping region

93 93

CHAPTER 11 CHAPTER 11 INEQUALITIES INEQUALITIES

Now N o w if system as asifif it were system of equations equations instead system of of inequalities, we would would if we we solved solved the system were a a system instead of a a system inequalities, we the intersection intersection point point of the get the the ttwo w o lines, case, happens be the solution solution with with the highest value value lines, which, which, in this case, happens to be the highest of x. As an find this this solution. Substituting the the first "equation" ”equation” into second, we get an exercise, exercise, let’s let's find solution. Substituting into the the second, we get 11 ‐ x ‐- 3 -‐ xx+ + 4 == -2x 2

:x‐6 -‐ 2 2xx++88=x4 -‐ 33xx = ‐- 114 .Xr

14 ‐ z ;::::: 44.66 .66 = z‐__3 3 - 14

this from from the the first equation). Therefore, (4.66, is the the solution solution At x x = 4.66, y = -‐4.66 4.66 + + 4 = -‐0.66 0.66 (we (we get get this equation). Therefore, (4.66, -‐O.66) 0.66) is with the the highest with There are no solutions in which x is 5, 6, or larger. highest value value of x. There are no solutions which larger.

While finding finding the While the intersection example may may have (hahal), these these points intersection point point in this this example have seemed seemed a bit bit pointless pointless (haha!), points can can be very be very important important in the context context of a given given situation, situation, such such as maximize profit as finding finding the the right right price price to maximize profit or figuring right amount materials for a construction construction project. figuring out out the the right amount of materials project.

EXAMPLE EXAMPLE4: 4: y

II

I

lII

IV

The following following system system of inequalities inequalities is graphed graphed in the xy-plane above.

2::‐- 33xx + 1 y2 y 22::22xx -‐ 3 Which quadrants quadrants contain Which solutions to the the system? contain solutions

A) Quadrants Quadrants I and and JI II

Quadrants I and and IV B) Quadrants

C) Quadrants Quadrants III I I I and C) and IV

D) Quadrants I, II, I I , and D) Quadrants I, and IV IV

First, graph equations, preferably your graphing graphing calculator. shade the the First, graph the the equations, preferably with with your calculator. Then Then shade the regions regions and and find the overlapping region. overlapping region . .1/

X

1

As you see, the overlapping region, which contains contains all region.. It has points in As you can see, the overlapping region, which all the the solutions, solutions, is is the the top top region It has points in quadrants I, II, I I , and quadrants and IV. Answer Answer ~.(D) .

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THE PANDA THE COLLEGE COLLEGE PANDA

Ecologists have have determined determined that the number number of EXAMPLE 5: Ecologists that the of frogs frogs y y must must be be greater greater than than or or equal equal to to tluee three times ecosystem to in a forest. In In addition, times the number number of snakes snakes x for a a healthy healthy ecosystem to be be maintained maintained in a particular particular forest. addition, the number of the number frogs and number of of frogs and the the number of snakes snakes must must sum sum to atwleast at.least 400. PART Which of the systems of of inequalities inequalities expresses conditions for for a healthy ecosystem? ecosystem? PART1: l! Which the following following systems expresses these these conditions a healthy

y 2~33xx y ‐- xx>400 >400

A) A)

y 2~ 33xx y‐x2400

y 2~33xx y+12400

B) B)

C) C)

D) D)

y ~ 3x 3,53:

y+x5400

PART2: IfIf the the forest forest currently currently has PART has a healthy healthy ecosystem, ecosystem, what what is the the minimum number of frogs in minimum possible possible number frogs in the forest? the forest? Part 1 Solution: Solution: The number number of frogs, y, must must be be at at least least three three times the number number of snakes, snakes, x. x. So, 50, y 2". 2 3x. 3x. The The

number of frogs frogs and and the number number of snakes snakes must soyy + number must sum sum to at least least 400, so + x 2". 2 400. Answer Answer ~-(C) . these types types of questions, the strategy the minimum minimum in the the graph graph of the Part 2 Solution: Solution: In ln these questions, the strategy is to look for the the inequalities. The minimum minimum (or maximum) maximum) will typically occur occur at at the intersection intersection point. point. To show show you you what what l1 inequalities. will typically mean, let's put the second inequality inequality in y = mx + mean, let’s first put the second = mx + b form.

33xx y 22". 00 y 22:‐ -x x++4 400 Now graph the the inequalities inequalities using using aa calculator Now we we can graph calculator.. y

The graph graph confirms confirms that that y, y, the the number number of at a minimum minimum at the intersection point. After all, the the The of frogs, frogs, is is at intersection point. After all, overlapping region (the top region) represents solutions and and the the intersection intersection point point is at the the bottom bottom overlapping region (the top region) represents all all possible possible solutions of this region, region, representing representing the solution solution with minimum number number of frogs. of with the minimum We can find find the the coordinates of that that intersection solving a a system system of equations We can coordinates of intersection point point by solving equations based based on on the ttwo wo lines. lines.

y= = 3x 3x = -‐xx + 400 y= Substituting the the first equation equation into into the second, second, Substituting = -‐xx + + 400 3x = : 400 4x = x = 100 X

So, 100 100 is is the the x-coordinate. x-coordinate. The Theyy‐coordinate -coordinate must beyy := 3x 3x = Given these these values, values, the So, must then then be = 3(100) 3(100) = : 300. Given the

I

intersection point point is is at at (100,300 and the minimum possible possible number number of frogs is j 300 when has a intersection (100,300)) and the minimum when the forest has health y ecosystem. healthy ecosystem. 95 95

CHAPTER 11 CHAPTER 11 INEQUALITIES INEQUALIT'IES

page 299. chapter start this chapter EXERCISE:Answers CHAPTER EXERCISE: Answers for this start on page

following allowed on the following calculator is allowed A calculator questions. questions. y

the to the solution to a solution following is a the following Which of the Which 14? 4x ‐- 14? 4 > 43: - x -‐ 4 inequality ‐x inequality

------.F------+

A)) -‐ 1 A

X

B) 2

C) 5 D) 8

Which of the following systems of inequalities could be the one graphed in the xy-plane above? A) y > 3 A )y>3

1 IfIf 3Zx‐4> ~x - 10, which of the following ~ x - 4 > Ex‐10,whichofthefollowing

yy > > xx

must be true? mustbetrue? B) y < 3 B) y < 3

A 4 A)) x 224

y xx

< ‐-224 C 4 C) ) x < D 4 > ‐-224 D)) x >

D D)) yy >> 3

y< ‐4(n If - 4(n -‐ 9), 9), what what is the the least least possible possible value value of n n? ?

C) 2 3 C) x X::::: 3 r < _3 X

'.S- 3

D) x S :S 3

xX 22 -‐33

work, Harry To get to work, Harry must must travel travel 8 miles miles by bus bus and 16 miles by train train everyday. and 16miles everyday . The The bus bus travels travels an average average speed speed of x miles at an miles per per hour hour and and the the train travels travels at train speed of y miles miles per at an an average average speed per hour . If lf Harry’s Harry's daily hour. daily commute commute never never takes takes more more than 1 hour, hour, which which of the than the following following inequalities inequalities represents the the possible represents possible average average speeds speeds of the the bus bus and train train during during the and the commute? commute? 8 16 A)§+ES1 A) -+-:S l X y x 16

y 8

x

y

B) -+-:S B>E+§g1 l X y

yy - +-‐< C) _ -2 x + 2

5 y g::;‐-Z2xx -‐ S Which of the the following following graphs the xy-plane xy-plane Which graphs in the could represent represent the the system system of inequalities inequalities above? above? could A) A)

140

~> 1100 >1,100 X + A) ‘x‐ + 10 10‐- Xx

A)

I

1, 100 B) 140x + + 80(10 -‐ x) > >1,100 140(10 -‐ x) > 1,100 C) 80x 80x + +140(10 1,100

B) B)

100 D) 80x + + 140(x 140(x -‐ 10) 10) > 1, 1,100

5 xx + y >>1l5 +aa

C) C)

y< 10 distance, assuming same distance, the same Amy have and Amy Jake and will minutes t w many minutes many i l l both both Jake have run r u n the 10 ? with Amy's distance rrun equate Jake's want to equate work with. variable t to work us aavariable gives us already gives problem already The problem The with. We Wewant Jake’s distance u n with Amy’s..

Jake’s distance: 60t Jake's distance: 20(t -‐ 10) distance: 120(10) Amy's Amy’s distance: 120(10) + + 20(t 10)

60t::120(10)4-20(t‐»10) = 120(10 ) + 20(t - 10) 60t 60t = : 11,200 , 2 o o+ + 20t 2 0 ; - 200 M = 11,000 bm) 40tn =

tf =z 25fi the same they will minutes, they After I2sIminutes, After will have have rrun u n the same distance. distance. the scientists. shared among be shared must be equipment must research equipment company, research a pharmaceutical EXAMPLE6: At a EXAMPLE pharmaceutical company, among the scientists. for freezer for one and one freezer scientists, and 3 scientists, every 3 for every centrifuge for one centrifuge scientists, one every 4 scientists, microscope for every one micromope There is one scientists many scientists company, how equipment at this company, 52 pieces there is a total scientists. If there every 2 every 2scientists. total of 52 pieces of research research equipment how many are there?

i,

. X

.

i,

, x

is 5, and centrifuges IS number of centrifuges the number is 1, the U1enumber Then the number of scientists. the number Let x be be the scientists. Then number of microscopes microscopes IS and . x

the IS ~ 5.. freezers is number of freezers the number X

X

x X

+ - = _ 552 2 -Z4 ++ -3 + 2

fractions, the fractions, both sides Multiply both Multiply sides by 12 12 to get get rrid i d of the

3 x+ 4 2 ~• 12 l2 4xx + 66xx z=552 3x 1 3 x=: 6624 24 13x

x:‑ 102

THE COLLEGE COLLEGE PANDA THE PANDA

EXAMPLE A group cost of renting a cabin cabin equally. equally. If each pays $130, EXAMPLE 7: 7:A group of friends friends wants wants to to split split the the cost renting a each friend friend pays they they will have $50 too too little. does it cost cost they will will have have $10 too too much. much. If each each friend friend pays pays $U0, $120, they will have little. How How much much does to rent rent the the -cabin? cabin? have two t w o unknowns problem.. We'll We’ll let let the number group be We have unknowns in this problem number of people people in the group be n and and the the cost of renting aa cabin cabin be be c. c. From From the the information information given, given, we we can come come up w o equations (make sure renting up with with ttwo equations (make sure you you see the the reasoning reasoning behind behind them): 13011 -‐ 10 = z cC 13011

+ 50 = cC 120n +

equation, 13011 130n represents represents the the total total amount amount the group 10 dollars dollars too much, In the first equation, group pays, pays, but but because because that’s that's 10 much, need to subtract subtract 10 10to arrive at the the cost cost of rent, rent, c. In the second second equation, total amount equation, 120n represents represents the total amount we need to arrive the group group pays, pays, but but this time time it's it’s 50 50 dollars dollars too little, little, so 50 to arrive at c. 6. Substituting from the so we we need need to add add 50 arrive at Substituting c from the into the second, the first equation equation into second, we we get get : 130n 130" -‐ 10 120n + 50 = -‐ 1lOn 0 n=: ‐ 660 0 = 6 n=

So So there there are are 6 friends friends in the group. group. And And

c:= C

13011‐10= 1 3 0· 6- 6- ‐ 10 1 0== 770 13011 - 10 = 130

renting the cabin cabin is is I-.770 1The cost of renting EXAMPLE 8: 8: Of the the 200 jellybeans jellybeans in in aa jar, 70% are are green rest are red. How How many many green jellybeans EXAMPLE green and and the the rest are red. green jellybeans must be be removed removed so so that 60% of of the the remaining remaining jellybeans jellybeans are green? must green? The answer answer is NOT N O T 20. You can't can’t just just take take 10% of the the green jellybeans away do that, green jellybeans away because because as as you you do that, the total total

:a

number jellybeans also also goes down. down. We We first find that that there jellybeans.. We x 200 = 140 green green jellybeans We need need to number of jellybeans there are are % x remove x of them so that that 60% of what's what’s left is green: green: remove them so

jellybeans left _ 60“/¼ green jellybeans green ------= 6O°oo totaljellybeans total jellybeans left ‘

140‐x_ 66 140 - X 200 - 110 2 0 0- ‐Xx _ 0 Cross multiplying, multiplying,

: 6(200 -‐ x) 10(140 -‐ x) = 1,400 10x := l,1,200 1, 400 -‐ lOx 200 -‐ 6x 200 = = 4x

=

x _‐_ 50 X 50

green jellybeans jellybeans need need to to be be removed removed.. This This type type of with percentages is very very common IsoIgreen of word word problem problem with percentages is common in in chemistry and and is typically known as asa ”mixture" problem problem.. chemistry typically known a "mixture"

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CHAPTER 12 WORD CHAPTER WORD PROBLEMS PROBLEMS

Our next next example example is the Our area / perimeter word word problem. problem. the classic area/perimeter EXAMPLE 9: A rectangle EXAMPLE rectangle has that is is 3 rectangle is is has a a width width that 3 inches inches shorter shorter than than its its length. length. It If the the area area of of the the rectangle square inches, inches, what 108 square what is the the perimeter, perimeter, in inches, inches, of of the the rectangle? rectangle?

Em I

If we we let the If be I, then then the the width w is/ is I -‐ 3. Since Since aa rectangle’s the length the length length be/, width w rectangle 's area area is is equal equal to to the length times times the the width, width, we can up the we can set up the following following equation: equation : : 108 llw w=

/(/ - 3) = 108 1(1‐ 12‐- 3131 ‐ 108 z= o 12 0 ( 1-‐ 112) 2 )(/( 1 ):o (/ + +9)9 = 0 Since the width is then /I -‐ 3 3 =z 12 12‐- 3 Finally, the the length length of a a rectangle rectangle has has to be be positive, positive, /I =: 12. The width is then 3 =z 9. 9. Finally, the

perimeter 21+ 210 = : 2(12) + + 2(9) =: I-.42 1. perimeter is 2/ + 2w Never forget forget that Never that the perimeter perimeter of a many a rectangle rectangle is twice twice the the length length plus plus twice twice the the width. width . I’ve I've seen seen too many students students just just add add the the length length and and the width width without without thinking thinking it through. through . EXAMPLE 10: When EXAMPLE When Alex Alex and Barry work work separately separately from from each paint a and Barry each other, other, Alex Alex can can paint a house house in in 6 6 days, days, and Barry can paint a house Barry can house in 12 12days. Assuming Assuming that they each work at a constant rate, how many days that they each work at a constant days will w i l l it it take Alex and Barry to paint a house if if they they work together? take.Alex and Barry together?

This is the the typical typical “work‐rate” "work-rate" problem This problem that involves two t w o individuals different rates. The general general that involves individuals who who work work at at different rates . The approach approach is to use use the r t , where where W W is is the the amount of work done, r is the overall rate at which the formula formula W = rt, amount of work done, r is the overall rate at which work being done, work is being done, and t is the time spent. The key thing to note is that the overall rate, r, can be found by and the time spent. The key thing note that overall rate, r, can be found by summing up summing up the the individual individual rates. rates. 11 Since Alex can days,, his his rate rate is 5 of a house 12 can paint paint a house house in 6 days house per per day. day. Since Barry Barry can can paint paint a a house house in in 12

1 days, his his rate per day. days, rate is ‐ of a a house house per 112 2

6

. . 1 _ 1 _ 2 1 _ 3 _ 1 Working + 12 =‐ 12 + 12 =‐ 12 = 4 of N o w we we can can use ‐ rt, rt, Working together, together, they can can pamt pamt 6 + of a a house house per per day. day. Now use W W=

6 12 12 12 12 4 where W = = 1 (i.e. (Le. 1 house) house) and z ~, 31, to find find the the time time it will will take paint one house.. where and r = take them them to to paint one house W= z rrtt 1 1l z= ‐- t 4

4 =: 1t Therefore, paint one one house. This answer makes sense sense because Therefore, it will take take Alex and and Barry Barry [±]days days to to paint house. Thjs answer makes because ifif Alex Alex can days by himself, then it should take less than 6 days if Barry is working can finish finish a a house house in 6 days himself, then it should take less than 6 days if Barry is workjng alongside alongside him. him.

104

THE COLLEGE COLLEGE PANDA PANDA

CHAPTER EXERCISE:Answers for this chapter start on page 301.

A calculator should N O T be should NOT be used used on the following following questions. questions. A rectangular rectangular monitor monitor has has a a length length of x x inches inches and and a a width width that that is one‐third one-third of its length. length. If the the perimeter 48 inches, what is the the perimeter of the the monitor monitor is 48 inches, what va lue of x? value x?

Which Which of the the following following represents represents the square square of the sum sum of x and the and y, y, decreased decreased by the the product product of and y y?? x and

A) x2 A) x2+ +y2 y 2 -‐ xy xy 2y 2 -‐ xy B) xx2y2 Jr};

C) (x+y)2‐ (x + y) 2 - ((xH+yy)) C) D) (x + y)2 -‐ xy D)

Susie salmon, weighing x Susie buys buys 2 pieces pieces of sa lmon, each each weighing pounds, pounds, and and 11 piece piece of trout, trout , weighing weighing y pounds, salmon pounds, where where x and and y are are integers. integers . The salmon cost $3.50 per per per pound pound and and the the trout trout cost cost $5 per pound. the fish was was $77, which pound . If the the total total cost cost of the which of the following be the value y? following could cou ld be the va lue of y?

On a a 100 cm cm ruler, On are drawn drawn at at 10, X, and ruler, lines lines are and The distance lines at X and and 98 cm. The distance between between the the lines times the the distance distance between 98 cm is three three times between the lines is the value of X ?? lines at X and and 10 cm. What What the value

A)) 4 A

B) 5 C) 6

D) 7

If 5 is added to the square root of x, the result is 9. What is the va lue of x + 2?

A 20% nickel alloy was was made nickel alloy made by by combining combining 22 grams nickel alloy with an grams of a a 35% nickel with 6 grams grams of an x% x % nickel nickel alloy. What What is the value value of xx ? ?

A grocery store sells tomatoes in boxes of 4 or 10. If Melanie buys x boxes of 4 and y boxes of 10, where x 2: 1 and y 2: 1, for a total of 60 tomatoes , what is one possible value of x ?

105

PROBLEMS WORD PROBLEMS CHAPTER CHAPTER 12 WORD

following allowed on the following A calculator is allowed questions. questions.

At a Hong Kong learning center,

i

of the

1 tak e studen ts take th e students debate, 2 of the tak e debate, studen ts take students

6

? If 8 + 5x is ’ twice ' x -‐ 5,, what h is ' the value off x ? 8 + 5x 15tw1ce x 5 w at 15the value 0 x - 6 A) ‐6 A

.. , The science . The take scrence. students take the students ~1 of the and § writing, and writlng, is what is math, what take math, 33 studen ts take math. If 33students take math. rest take rest learning the at students of number total number students at the learning the total the 7 center? center.

) -3 B) ‘3 B) C) _§7 3 C)

A 0 A)) 660 B) 66 B) 66

D) D) -‐22

C)) 7722 C

_

D

)

7

D) 78 8

the what is the of n, what the same is the of 68 is 75% of IfIf 75°/o same as 85% of

vvalue a l u eofdn? "?

_ has 44 Jason has cards, and 20 footba ll cards, has 20football Ian has Ian and Jason 44 that such that trade such agree to trade They agree cards . They baseball cards. baseball card every card cards for every 2 baseball cards Ian 2baseball gives Ian Jason gives Jason trades suc h trades many such how many After how Jason. After Ian gives to Jason. Ian gives number of equa l number an equal have an each have and Jason Ian and will Jason each will Ian cards? cards?

|

|

_

A

)

9 A) 9

B 0 B)) 110 C) 11 C) 11

games . their first 15 The o n exactly 15 games. exactly 4 of their won Pirates w The Pirates won and won games and remaining games They played N remaining then played They then the all the half of all exactly half won all o n exactly they w them. If they all of them. value of N ?? the value what is the played, what games they played, games they

D) 12 D) 12

the number x, the the number times the from 3 times IfIf 3 is subtracted subtracted from added to when 8 is added result when the result What is the result result is 21. What half of x ? A) 1

|

|

_

B

B) 5

)

5

C) 8

12 D) 12 D)

the same with the Alice start with same number number of Julie start and Julie Alice and pens her of 16 gives pens. After Alice gives 16of her pens to Julie, Julie, Alice After pens. Alice as pens Julie w o times many pens asAlice as many times as has ttwo then has Julie then the have at the Alice have did Alice pens did does. o w many many pens How does. H start? start?

106

PANDA COLLEGE PANDA THE COLLEGE

i i

books on on aa the books and g of the Mark w n %and own Kevin o and Kevin Mark and

than less than dollars less a tie is k dollars price of a the price store, the a store, At a costs shirt a If shirt. a of price the price a shirt. If a shirt costs $40 times the three times three value of k? the value what is the costs $30, what tie costs and a tie and k?

the rest of the owns the Lori owns respectively . Lori shelf, respectively. shelf, the rest than books than Mark, more books Kevin oowns If Kevin books . If books. w n s 9 more Mark, own? Lori own? does Lori books does many books how many how

has rectangle has a rectangle shape of a the shape board in the wooden board A wooden the area of the the area If the width. If its width. twice its that is twice a length that a length length, in feet, what square feet, board is 128 square board what is the the length, board? the board? feet, of the feet,

grand its grand celebrate its coupons to celebrate out coupons gave out A bakery bakery gave either $1, $3, opening. Each coupon coupon was was worth worth either opening. Each out given out were given $1 coupons were many $1coupons as many Twice as or $5. Twice coupons $3 many as times 3 and 3 times as many $3 coupons coupons, and as$3 as $3 coupons, value of The total coupons. The as $5 coupons. given oout were u t as$5 total value were given out was coupons given all given out was $360. How H o w many many the coupons all the out? given out? were given coupons were $3 coupons $3 A)) 440 A 0 B) 45 45

C) 48 C) 48 D 4 D)) 554

has seas hells . Bob has collect seashells. all collect Carl all and Carl Alex, Alex, Bob, and three Alex ha as Carl. many seashells half seashells as Carl. Alex hass three as many half as If Alex and as Bob. If times and Bob seashells as many seashells as many times as seashells many seashells how many seashells, how have 60 seashells, together together have have? Carl have? does does Carl

A water tank is connected to two pipes, Pipe A and Pipe B. It takes 4 hours to fill the tank when only Pipe A is in use, and it takes 6 hours to fill the tank when only Pipe Bis in use . If it takes 111 minutes to fill the tank when both Pipe A and Pipe Bare in use, what is the value of m ?

A 5 A)) 115

B) 20 20 C 0 C)) 330

D D)) 4400

107

CHAPTER 12 WORD CHAPTER WORD PROBLEMS PROBLEMS

Yoona runs steady rate yard per second.. runs at a steady rate of 1 yard per second Jessica runs Jessica gives runs 4 times times as as fast. IfIf Jessica gives Yoona head start start of 30 yards a head yards in a race, race, how how many many yards must must Jessica Jessica rrun yards u n to catch catch up to Yoona?

Terry is hired hired to pave pave a a parking parking lot lot and and finishes finishes 1 3 of the hired to the parking parking lot lot before before Andy Andy is is hired to work work

3

alongside him. alongside him . They They each each work work at at a a constant constant rate , but but Terry works rate, works twice as as fast as as Andy Andy does. does.

x

1 + i)1 ) 2 can be used to The equation equation 9 9 (i x = 2 can ( + 3 be used to 2 the total total number number of days find the days x it would would have have taken Terry to pave pave the taken lot by by the entire entire parking parking lot himself. Which Which of the the following himself. the best following is the best interpretation of the the number number 9 in the interpretation the equation? equation?

A) The The number number of days days it would would have have taken taken Terry and and Andy Andy to pave pave the parking the entire entire parking they had had worked worked together together from from the lot ifif they the start. start. number of days days it will will take take Terry and B) The number and Andy to pave pave the the remainder remainder of the the parking Andy parking working together. together . lot working The number number of days days it would would take take Andy Andy to C) The pave the the remainder remainder of the the parking parking lot pave lot if if he he were working working alone. alone . were The number number of days days it would would take take Terry D) The Terry to pave the the remainder remainder of the the parking parking lot he pave lot ifif he were working working alone. were alone.

Nicky owns a house that has a patio in the shape of a square . She decides to renovate the patio by increasing its length by 4 feet and decreasing its width by 5 feet. If the area of the renovated patio is 90 square feet, what was the original area of the patio, in square feet?

108

13

Minim Minimum Maximum um & Maxim um Word Word Proble Problems ms Minimum Minimum and and maximum word problems problems require require aa bit bit of maximum word of logic and and an an understanding understanding of of rates rates and and inequalities inequalities (chap (chapters and 11). One One of the most most common common issues students have is that they’re unsure of whether round ters 4 and issues studen ts have is that they 're unsure of whether to to round up or down. down. The examples examples in this chapter will address this issue and illustrate the strategies you’ll need to this chapter address issue and illustrate strategies you' ll need to solve solve these types types of problems. problems.

EXAMPLE 1: Corinne graphic designer designer who who earns logo she designs.. What What is is the the EXAMPLE Corinne is aa graphic earns $275 for for every every logo she designs minimum number number of logos logos she would would have have to design design to earn least $4,000 minimum earn at at least $4,000 ? To earn Corinne would would have have to design design at least least 4, To T earn at least least $4,000, $4,000, Corinne \~

0

_ x 14.5 14.3 logos. logos. That's That’s 14 14 logos logos and and half half aa ::::::

logo. But becau because it’s implied implied that that a fraction of a logo logo cannot up to to se it's cannot be be designed designed and and sold, sold, we have have to round round up ii logos. OI) logos. When When a whole whole number is implied, implied, the the minimum m i n i m u m generally that we we round up. number answer answer is generally requires requires that round up. EXAMPLE EXAMPLE 2: A pallet pallet truck truck can can move move up up to 3 tons single trip. be used used to to move move tons in a single trip. If If the the truck truck is is to to be 320-pound 320-pound pallets, pallets, what what is the the maximum maximum number number of whole pallets the truck can can move move in in aa single single trip? trip? whole pallets the truck (1 ton pounds) ton = 2,000 pounds)

A) 6 A)6

B) 18

C) 19

D) 106

6, 000 6,000 : 18.75 pallets. pallets. However, However, = 320 the question question specifically specifically states states whole whole pallets, pallets, so sowe have to E pallets pallets.. If rounded up, up, the we have to round round down down to to DI] If we we rounded the weight would weight would be be above above what what the truck truck can handle. handle.

Since 3 tons is equivalent equivalent to 3 x 2, 000 =: 6,000 pounds, pounds, the truck can move move 2,000

When number answer generally requires When a whole w h o l e number a n s w e r is implied, implied, the maximum maximum generally requires that we we round round down. down.

109

MAXIMUM WORD MINIMUM & MAXIMUM CHAPTER 13 MINIMUM CHAPTER WORD PROBLEMS PROBLEMS

what yogurt, what greek yogurt, of greek 6 cups and 6 flour and of flour cups of 8 cups flatbread can tray of flatbread one tray EXAMPLE 3: H If one can be be made made from from 8 cups of and 100 of flour cups of from 150 made from be made can be that can flatbread that whole trays number of whole maximum number the maximum is is‘the trays of flatbread 150 cups flour and 100 cups cups yogurt? gteEtkyogurt? of greek other, used up be used will be yogu rt) will greek yogurt) resources (either the resources questions , one In these one of the (either flour flour or greek up before before the the other, types of questions, these types In these approach to way best the Therefore, produced. be can that amount the limit resource will and that will limit the amount that can be produced. Therefore, the best way to approach these that resource and resource separately. each resource consider each questions is to consider questions separately. 10

the consider the only consider we only If we made . If be made. can be flatbread can tra ys of flatbread l~O = 18.75 trays requirement, % the flour consider the only consider we only IfIf we flour requirement,

100

be produced can be that can amount that the amount Since the made . Since be made. can be flatbread can trays of flatbread l~O ~ requirement, ‐6‐ yogurt requirement, greek yogurt greek 2 16.7 trays produced limiting a limiting is a yogurt is greek yogurt the greek flour, the the flour, produced from can be that can amount that the amount than the less than from greek yogurt yogurt is less beproduced from the the greek from the the result, a As . flatbread of trays of flatbread. As a result, the limited to 16.7 trays flour, so use up enough of it to use isn't enough factor. up the the flour, so we're we’re limited There isn’t factor . There

when finding round down we round that we Remember that made is [!I]. be made can be that can trays that number of trays whole number maximum -. Remember down when finding the the maximum whole maximum. maximum. EXAMPLE 4: EXAMPLE4:

= 18tw 18tw + + 1,050 1,050 C=

producing of producing dollars, of in dollars, C, in total cost the total calculate the equation above uses the appliance manufacturer An appliance manufacturer uses the equation above to calculate cost C, more than no more spend no can spend manufacturer can the manufacturer that ea-ch toasters that shipment of t toasters a shipment a each weigh weigh w to pounds. pounds. If If the than $21,000 $21,(X)0 the what is pounds, what 6 pounds, be 6 will toaster w each toaster and the toasters, and shipment of toasters, producing producing the the next next shipment the weight weight of each i l l be is the shipment? next the for produced be can that toasters of number maximum maximum number toasters that can be produced the next shipment? up an setting up question by this question solve this Let's solve Let’s by setting an inequality: inequality: c ~ 5 21,000 21,000 C

21,000 ::S:21,000 18110 + + 1,050 1,050 5 lBtw 18t(6) ::S: 5 19,950 19,950 18t(6) 10815 19,950 ::S:19,950 108t 184.72 ::S:184.72 t5 produced be produced number that maximum number the maximum numbers , the whole numbers, produced in whole are produced toasters are that toasters implied that it's implied Since it’s Since that can can be

I184 1. for the shipment is -. next shipment the next

110

THE COLLEGE PANDA THE COLLEGE PANDA

EXAMPLE deck of48cards of 48 cards consists EXAMPLE 5: 5: A Adeck consists of ofonlynedcards Ifthenumberofredcards only red cards andblackcards. and black cards. If the number of red cards is less than twice twice the the number less than number of black what is the in the black cards, cards, what the minimum minimum possible possible number number of of black black cards cards in the

deck? deck? Let r be be the the number Let number of red red cards cards and and b be the number variables, we can set be the number of black black cards. cards. Using Using these these variables, we can set up up a a system that consists consists of an system that an equation and an an inequality. equation and inequality.

r+b= : 448 8 b r >0 B) B) mb mb < 00

(1, 2), what what is the the (1,~),

C) mb = 0 mb =

value of gg(‐1) value (- 1) ? ?

D) mb : 1 mb =

The line y = -‐ 22xx -‐ 22 is perpendicular perpendicular to line/. line I. If If these two lines have the same y-intercept, which of the following could be the equation of line / ?

A A)) yy=z ‐- Z 2xx-‐ z2 B 2 x- ‐ 2 B) ) yy== 2x

A calculator is allowed allowed on the following following questions. questions.

1 C) C) .v yz‐ix‐Z =-- 2 x - 2

1 D)y=%x‐2 D) y = - x - 2 2

y m

The slope slope of line line/ I is ~%and and its its y‐intercept y-intercept is 3.

What What is the equation equation of the the line line perpendicular perpendicular to line (1,5) line I that that goes goes through through (1, 5) ??

A = ‐ 22xx + 3 A)) yy =B = ‐ 2x h+ B) ) yy =+7

What m in the figure What is the slope slope of the the line line m the figure above? above?

A) A) - 2

1 11 11 C) C) y y:_§x+7 = - 2x + 2

1 B B) ) ‐ ‑

1 9 D) D) yy‐§x+§ = 2x + 2

2

1 C) C) 4

1 D) 2

124

THE COLLEGE PANDA PANDA THE COLLEGE

2

A line line with with aa slope passes through slope of g passes through the the

In the xy-plane, the line with equation 1 . ax ‐ 531 := 8, where constant, passes where a 15a is a constant, passes

3

-1y

points (1, 4) and points (1,4) and (x, (x, 10). 10). What What is the the value value of x? x? A) A) 4

through point (2,6). What is the through the the point (2, 6). What the x-coordinate of the x-coordinate the x-intercept x-intercept of the the line? line?

B) 6 C) C) 8 D) D ) 1100

D Day ay

Monday Monday Thursday Thursday ”‑ Friday Friday Saturday Saturday

Average speed, speed , 5 s (miles per per hour) hour)

Number of Number calories burned, burned , c

7.2

616

6.8

584

7.9

672

8.5

720

a

y = bx + c d +cC y = ‐- xX + y e The lines in the The equations equations of two two perpendicular perpendicular lines the xy-plane are shown above, and e a, b, c, d, and e xy-plane are shown above, where where a,b,c,d,

On certain certain days week, Elaine an days of the the week, Elaine runs r u n s for an hour on treadmill. For each day that that she she ran ran in each day hour on aa treadmill. the week, the the table shows the the average the last last week, table above above shows average speeds at which speed sat which she she ran, ran, in miles miles per per hour, hour, and and the number number of calories she burned burned during during the the the calories c she run.. If relationship between between cc ands and s can can be be run If the the relationship modeled function, which the modeled by aa linear linear function, which of the following functions best best models models the the following functions relationship? relationship?

a

following must be true? following must be true?

d A) - < - 1 A)‘1 0. 0. By thiss makes sense. Because Because of of the the square square in in the the real numbers. numbers. Put mathematically, f(x) By the the way, way, thi makes sense.

denominator (x) = = xm2 denominator of of ff(x)

_

l~x

+ 25

= fi, (x

you always positive output output for any va value in the the ~ 5 ) 2 , you alwa ys get get a a positive for any lue of of xx in

domain. domain . Let’s the domain domain,, start start with with all all real and then the values values of xx for for which Let's summarize. summarize. To find the real numbers numbers and then exclude exclude the which the function function is invalid invalid or undefined. For examp example, 2 0 take the the the undefined . For le, the the domain domain of y := fl.jx is is xx ~ 0 because because we we can't can't take square negative numbers numbers.. square root root of negative

To find the graph the on your your calculator calculator and u t the possible values values of of y, y, taking taking note note of of the range, range , graph the function function on and figure figure o out the possible any asymptotes. any horizontal horizontal asymptotes.

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CHAPTER 16 CHAPTER 16 FUNCTIONS FUNCTIONS

EXAMPLE 3:I f f ( x -‐ 1) 1) = and g(x) == xx + 3,what is the value (2))? off(g(2)) ? EXAMPLE3:If/(x = 6x 6xandg(x) + 3,whatisthevalueof/(g Whenever you of other other functions), functions), start Whenever you see composite composite functions functions (functions (functions of star t from from the the inside inside and and work work your your way way out. out. First, First, g( 2) = 2 + 3 = 5 Now we have have to (5). Now we to figure figure out out the the value value of of f/(5). Well, we into J(x f ( x -‐ 1) 1) == 6x 6x to to get getf(5) = 6(6) = I-.36 I. we can can plug plug in in x x := 6 6 into f (5) = 6(6) =

EXAMPLE byf(x) EXAMPLE 4: 4: Functions Functions/ f and and g g are are defined defined by f(x) the value value of k ?7

i·

= xx+1 = §.1ff(g(f(k))) = 10, what is is = + 1 and and g(x) g(x) = If f(g(f(k))) = 10, what

Again, our way way out: out: Again, we we start start from from the the inside inside and and work work our

f ((k) k )=: k + 11

= k+1 g(k+1):k%1 2

g(k + 1)

1c;1)

= k ;1+

1

Finally, Finally, k+l -I%1+1210 2- + 1 = 10 k+1 = k%1:9 9 2

8 kk + 1l 2=118

= [ill k=E

k

As we've returns an and output output pairs allow we 've mentioned, mentioned, a function function takes takes an an input input and and returns an output. output. Well, Well, these these input input and pairs allow us function asa set of points in the xy‐plane, with the input as x and the output asy. In us to graph graph any any function as a set of points in the xy-plane, with the input as x and the output as y. 1n fact, fact, 2 y = xx2+ the same 1. Both and yy are are the + 11is is the same asf as f (x) = x2+ x2 + 1. Both ff(x) (x) and the same same thing‐they’re thing-they're used used to to denote denote the the output. output. The reason we use with the The only only reason use y is that that it’s it's consistent consistent with the y‐axis y-axis being being the the y‐axis. y-axis.

Anytime (x) is used of it it as as the (x) > > 0, Anytime ff(x) used in a graphing graphing question, question, think think of they.y. 50 So ifif a a question question states states that that fJ(x) 0, all allyy values graph is is always above the x-axis. It’s extremely important that you learn to think values are are positive positive and and the graph always above the x-axis. It's extremely important that you learn to think of points inputss and and outputs points on on a a graph graph as as the the input outputs of of a a function. function .

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THE PANDA THE COLLEGE COLLEGE PANDA

EXAMPLES: EXAMPLE 5: y ;

I/\ \

J

- 1-

'O

I I

-·

\

\ I

\

X

\

\ I

\

The f (x) is is shown shown in in the xy-plane above. above‘ For what value value of (x) at The graph graph of of J(x) the xy-plane For what of x xisis fJ(x) at its its maximum? maximum?

Again, (x) as as they the y.. So So we’re graph with with the Again, when when it it comes comes to to graphs, graphs, think think of offf (x) we're looking looking for for the the point point on on the the graph the highest y‐value, the ”peak”" of That point point is x-value is .. highest y-value, the "peak of the the graph. graph. That is (5, (5, 4), 4), and and so so the the x-value is I}].

EXAMPLE the function function with with equation = ax a):22 + point (1, 2), what what is EXAMPLE6: If the equation yy = + 33 crosses crosses the the point (1,2), is the the value value of of a a?? Remember - a point Remember‐a is just an output, output, an an xx and y. Because 2) is is a on the the graph point is just an an input input and and an and a a y. Because (1, (1, 2) a point point on graph of of the the function, function, we we can can plug plug in 1 for x and and 22 for y.

2 2 =:a(1)2 a(1 )2++3 3 2 = aa + 3

a z j 2+ EXAMPLE EXAMPLE 7: If the function y = = xx2 + 2x -‐ 4 contains the point and m m> what is is the value the function contains the point (m,2m) (m, 2m) and > 0, 0, what the value of m? ofm?

It's important important n not get intimidated It's o t to get by all the variables. variables. The gives us us aa point point on on the graph, so so let's let's intimidated by all the The question question gives the graph,

plug plug it it in. in .

: x2 yy = x2 + +2x 2x -‐ 4 4 2m = m2 + 2m 2m-‐ 44 2m m2 + 00 = m2 m 2 -‐ 4 4

From here From we can can see that :l:2. The question question states states that : l.@]. here we that m = ± 2. The that m > 0, 0, so so m m=

The zeros, zeros, roots, and x-intercepts The function are are all for the the same thing‐the values x-inter~epts of aa function all just just different different terms terms for same thing - the values of x that that make make ff(x) (x) = = 00 (or y = = 0). O). Graphically, Graphically, they values of crosses they refer refer to to the the values of x x where where the the function function crosses the x-axis. the

135

FUNCTIONS 16 FUNCTIONS CHAPTER 16 CHAPTER

EXAMPLES: EXAMPLE8:

y ~

I

I

I

,

--

I

I

I I I

'

. "" .."

.,_:. I I

'\

--

'\ I I

'\

'

X

'

I

I I I

" ...../

I

I

The graph off (x)

= x3 - 2x2- Sx + 6 is shown in the xy-plane

above.

does the zeros does distinct zeros PART 1: How How many many distinct the function function ff have? have? value the value be the could be following could the following which of the solution, which that J(x) such that constant such If k is a constant PART 2: If f (x) = = k has has 1 solution, of kk??

-3 A) ‐3 A)

B) 1

C) 5

0)99 D)

has so ff has times, so three times, x-axis three the x-axis crosses the graph crosses The graph Solution : The Part 1 Solution: Part and 3. are -‐ 22,, 1, and zeros are these zeros that these see that see

can we can the graph, distinct zeros ffidistinct zeros.. From From the graph, we

Read explanation. Read the explanation. during the lost during you feel lost panic ifif you so don't involved, so quite involved, question is quite This question So lution: This Part 2 Solution: don’t panic sense make to able be you'll promise I . confusing were that bits the to back go then go back the bits that were confusing. promise you'll be able make sense of and then through and the way all the all way through everything. everything. always we always input, we the input, matter the function. No matter just a function. realize that this question, To truly understand question, first realize that a constant constant is just understand this does What does say k = -‐ 33.. What let's say So let’s y = k or g(x) = k. So can write we can question, we this question, output. In this same output. the same get the get write it as asy at -‐ 33!! line at horizontal line like? A horizontal y z= ‐3 look like? - 3 look

y

intersection points the intersection referring to the merely referring it's merely = k, it’s so lutions to fJ (x) = the solutions asks for the question asks a question Now points of f (x) when a Now when (x) J other, each other, f(x) = equal to each functions equal sets ttwo question sets a question general , if a k. In general, line y = horizontal line and = k. w o functions = g(x), and and the horizontal and the intersection points the intersection only at the it's only all, it’s After all, points. After intersection points. the intersection referring to the it's referring solutions, it’s the solutions, about the asks points you about asks you constant a constant be a happens to be just happens g(x) just case, g(x) particular case, this particular functions. In this both functions. that same for both the same value of y is the the value that the function, : k. g(x ) = function , g(x) there above, there shown above, as shown - 3 as So ifif k = points. 50 intersection points. the number The equivalent to the number of intersection : ‐3 solutions is equivalent number of solutions The number the are the themselves are solutions themselves The solutions points. The must intersection points. the 3 intersection represented by the as represented solutions to fJ (x) = -‐ 33,, as be 3 solutions must be and 2.6. 2.2, 1.6, and be -‐2.2, them to be x-values estimate them can estimate points . We can those points. x-values of those

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THE COLLEGE COLLEGE PANDA THE PANDA

Getting back we have choose a Getting back to the the original original problem, problem, we have to to choose ak k such such that that there there is is only only one one solution. solution. Now Now we’re we're thinking backwards. thinking Instead of being constant, we have to to choose choose it. it. Where might we place place aa backwards. Instead being given given the the constant, we have Where might we horizontal there’s only only one intersection point? o t at we just just showed how horizontal line line so so that that there's one intersection point? Certainly Certainly nnot at ‐3 - 3 because because we showed how that would would result that result in 3 solutions. solutions . Well, looking looking back one just below -‐ 44.. Horizontal Horizontal lines at these back at at the the graph, graph, we we could could place place one just above above 88 or or just just below lines at these values w i t h fJ (x) just just once. Looking at is the values would would intersect intersect with once. Looking at the the answer answer choices, choices, 9 9 is the only only one one that that meets meets our our condition. Answer Answer ~(D) .condition. Let’s take a revisit part part 1. 1. In In part 1, we we found number of intersection points between fJ (x) and Let' stake a moment moment to revisit part 1, found the the number of intersection points between (x) and But realize realize that that the just the the horizontal horizontal line y = 0. In counting the number of intersection the the x-axis. x-axis . But the x-axis is is just line y = 0. In counting the number of intersection points = 0, really doing doing is is finding the number number of points between between f (x) and and the the horizontal horizontal line line yy = 0, what what you you were were really finding the of solutions 0. solutions to f (x) = =

If you you didn’t in this this example the first first time through didn't grasp grasp everything everything in example the time through, through, it’s it's ok. ok. Take Take your your time time and and go go through it again, making sure throw quite questions at at again, making sure you you fully fully understand understand each each of of the the concepts. concepts . The The SAT SAT will will throw quite aa few few questions you as well well as as the the solutions solutions to (x) = = g(x). you related related to the the zeros zeros of of functions functions as to fJ (x) g(x) . Hopefully by by now, you’re starting to see constants ashorizontal lines. 5, that means Hopefully now , you're starting to see constants as horizontal lines. 50 So for for instance, instance, if if fJ (x) (x) > > 5, that means the is above z 5. in this i l l help help you you on on a the entire entire graph graph of offf is above the the horizontal horizontal line line y = 5. Thinking Thinking of of constants constants in this way way w will a lot of SAT graph lot graph questions. question s. 3 + 2x2 EXAMPLE 9: Which of following could could be be the the graph EXAMPLE 9: Which of the the following graph of of y y = xx3+ 2x 2 + + xx ++ 11??

B) m

A) ~

C) q

#‘x $3: %x yy

y

y

D) yy

x

Although Although the the given function looks complicated and you might given function looks complicated and you might be be tempted tempted to to graph graph itit on on your your calculator, calculator, this this is the find a be on on the graph and eliminate eliminate the easiest easiest question que stion ever! ever! A Alll l you you have have to to do do is is find a point point that’s that's certain certain to to be the graph and the So what’s an easy easy point and test? test? the graphs graphs that that don’t don 't have have that that point. point. So what 's an point to to find find and

Plug to get = 1. 1. N o w which which graphs graphs contain contain the Plug in x x z = 00 to get y = Now the coordinate coordinate (0, (0, 1)? 1)? Only Only graph graph ~-.

-

By are particularly particularly good good for to use By the the way, way, numbers numbers like like 00 and and 11 are for finding finding ”easy” "easy" points points to use for for this this strategy. strategy.

137

CHAPTER 16 CHAPTER 16 FUNCTIONS FUNCTIONS

Function Transform Function Transformations ations Function transformations Function transformations are are changes changes we we can make make to to the the equation equation of of a a function function to to ”transform” "transform" its its graph graph in in specific ways. The transformations you might encounter on the SAT are reflection across the x-axis, ways . transformations you might encounter on the SAT are reflection across the x-axis, vertical verticaJ shift, horizontal shift, and We’ll cover first three shift, horizontal shift, and absolute absolute value, value. We'll cover the the first three here here and and discuss discuss absolute absolute value value transformations transformations in the absolute absolute value value chapter. chapter. Let's start Let’s with the (x) = z x2 + 2x, start with the example example function function fJ(x) x2 + 2x, whose whose graph graph looks looks like like y

graph of fJ(x) To reflect the graph (x) across down),, multiply ( x ) by by -‐ 11.. The The resulting across the x-axis x-ax is (flip (flip itit upside upside down) multiply ff(x) resulting equation, equation, 2 - 2x, produces y = f(x) = x the reflected 312 ~f (x) z ‐x2 ‐ produces reflected graph. graph. yy = = -‐ f(x) fl fl y ·. I

To shift shift the graph up, add To graph of fJ(x) ( x ) up, add a constant ( x ) . For example, ( x ) ++ 22 =z x2 + 2x constant to ff(x). example, y 2= fJ(x) x2 + 2x + + 22 produces produces aa graph that that is 2 units graph (x). units above above the graph graph of of ff(x). shift the graph graph of fJ(x) down, subtract To shift (x) down, (x). For (x) -‐ 2 + 2x subtract a constant constant from ff(x). For example, example, yy = = fJ(x) 2 z= x2 x2 + 2x -‐ 2 2 produce s a a graph produces graph of f ( x). graph that that is 2 units units below below the the graph of J(x).

yy = = f(x fl n) ++22

y = f(x) - 2

y

y

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COLLEGE PANDA THE COLLEGE PANDA

To shift x ) to the the left by a [1units, units, substitute substitute x + + a in = fJ((xx + + 1) 1) = = shift the the graph graph of fJ((x in for for x. x. For For example, example, yy = 1)22 + 2(x + 1) 1) produces produces a graph that that is unit to left of the graph of f(x). (x + 1) a graph is 11 unit to the the left of the graph of J (x). To shift x ) to the substitute x -‐ a shift the the graph graph o offff ((x) the right right by a11units, units, substitute a in in for x. x . For For example, example, yy = = fJ((xx -‐ 1) =z + 2(x -‐ 1) 1) produces is 11 uunit n i t to off(x). (x -‐ 11)2 )2 + produces a a graph graph that that is to the the right right of of the the graph graph of J (x).

y = f (x + l )

y

= J (x -

y

1)

y

Here's Here’s a trick that use to make make sense that I like to use sense of horizontal horizontal shifts: shifts : For horizontal For horizontal shifts, shifts, find find out out what what value value of of xx makes makes the equal to the substituted substituted expression expression equal to 0. 0. This This value value will will tell tell you horizontal shift shift is. is. For For instance, instance, when when we we have f (x ‐ 1), what value of x makes 1 you what what the the horizontal have J(x - 1), what value of x makes x x -‐ 1 equalto 0?x = 1. Sof(x ‐ 1)is1 unit tothe rightoff(x). What aboutf(x+4)? Well,x = ‐4makesx+4 equal to 0? x = 1. So J(x - 1) is 1 unit to the right off (x ). What about J(x + 4)? Well, x = - 4 makes x + 4 equal to equal to 0. Since value is negative, graph is is shifted left. And And what what about f(3x -‐ 2)? Since the the value negative, the the graph shifted 4 units units to to the the left. about /(3x 2)? What's5 the the horizontal horizontal shift What’ relation to to f(x)? f (x)? A value value of x = = ~ 2 makes equal to the horizontal horizontal shift in relation makes 3x -‐ 22 equal to 0,so 0, so the 2 shift is is ~3 units units to right. shift to the the right.

Note that horizontal horizontal and Note that referred to astranslations. the graph of a and vertical vertical shifts shifts are are commonly commonly referred as translations. And And the graph of a transformed transformed hmction is often function often called called an an image image of the graph of the the original original function. the graph function . EXAMPLE EXAMPLE 10: In the the function function g units to to the the left left the xy‐plane, xy-plane, the the graph graph of the g is the the graph graph of of f/ translated translated 55 units and 33 units units downward. downward. If and the function (x) = (x -‐ 3) 3)22 + + 1, 1, which which of of the the following lithe function f is is defined defined by fJ(x) following defines defines g(x)? g(x)?

A)g(x) A)g(x)=(x‐8)2‐2 = (x - 8)2 - 2

B)g(x)=(x+2)2‐2 B)g(x) = (x + 2)2 - 2

C)g(x)=(x‐8)2+4 C)g(x) = (x - 8)2 + 4

D)g(x)=(x+2)2+4 D)g(x) = (x+2) 2 +4

A translation units downward downward means translation of 5 units units to the the left and and 3 units means that that

g( J (x + 5) -‐33 3 0x )) = =f(-Y+5)

: ( (x ( x + 55)) -‐ 33)) 22+ + 11-_ 33 == (x+2)2~2 = (x + 2) 2 - 2 Answer Answer ~(B) . To summarize, summarize, for a function f (x),, To function J(x)

o (x)) results in a x-axis • -‐ Jf (x results in a reflection reflection across across the the x-axis 0 x) + + a results units; Jf(.r) of a a units units • fJ((x results in in an an upward upward shift shift of a units; (x) ‐- 11 a results results in in a a downward downward shift shift of 0 + a) results shift of a units horizontal shift shift of a units • f (x + results in a horizontal horizontal shift units to to the the left; left; f (x ‐- (1) a) results results in in a a horizontal of a w1its to to the right the right

139

CHAPTER CHAPTER 16 16 FUNCTIONS FUNCTIONS

CHAPTEREXERCISE:Answers for this chapter start on page 311.

A calculator should NOT be used on the following questions. y

X

y

0

20

1

21

3

29

The table table above above displays displays severa The severall points points on on the the graph of the the function function fin graph f in the the xy-plane. xy-plane. Which Which of the following could could bef be J(x)? the following (x) ?

A) f (x)) = 20x A) J(x B) fJ(x (x)) z= xx ++20 8) 20 C) fj (x) (X) = = xX -‐ 20 20 C) 2 D) D) f(x) J(x )==x2 x ++20 20

The graph graph of the the function function f is shown the The shown in the xy-plane above. If J(a) = f(3), which xy-plane above. If f (a) f (3), which of the the following could could be be the following the value value of of aa ??

A)) -‐ 4 A 8) ) -‐ 3 B C)) -‐ 2 C

D) 1

y J(x)

y

In the portion of the xy-plane shown above, for how many values of x does J(x ) = g(x)? A) None B) One

The function f is graphed in the xy-plane above . For how many values of x does J(x) = 3?

C) Two

D) Three

A) Two 8) Three C) Four D) Five

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For which of the following functions is it true that J(-3) ==f(3) J(3)?? thatf(‐3)

IfIf fJ(x) (x) = values = x2, x 2 , for for which which of of the the following following va lues of of

c isf is ((c) c) < cc??

A) f J(x) M m ==-X2;

1 A) 1 A)E 2

x3

B 8) ) 1

8) J(x) = B)fu%=§ 3

3 C) ~ C)z 2

C) C) f(x) J(x) =3x2+1 = 3x 2 + 1 D) f(x) 0) f(x) = xx + + 22

D) ) 2 D

the graph graph of the IfIf the the function has x-intercepts function fJ has x-intercepts at at -‐33 and and 2, and 12, which which of of the and ay-intercept a y-intercept at at 12, the following could following could define define f ? ?

The hmction f is defined by J(x) = 3x + 2 and the function function g is defined the by g(x) defined by g(x) = f(2x) -‐ 1. = J(2x) l. What is the What the value value of of g( 10) lO) ? ?

A) fJ(x) (x + 3)2(x -‐ 2) A) ( x )‐=‐ (x (+3)2110 0 f(O) D) D ) ff(O) ( 0 )= ‐‐ 10 10

A) g(l) B) g(3)

C) g(5)

D) g(B)

y y

--

Y = g(x)

-

1 +-----::Q +----------1----+

X

1

'

.

•--~ I

!

•

····-

I ..

The graph of the function g is shown in the xy-plane above, and the function f (not shown) is defined by f (x) = x3 . If g is defined by g(x) (x + a) + b, where are constants, g(x) := fJ(x +a)+ where a and and b bare constants, what what is the the value value of a a+ + bb??

The graph graph of f(x) The f (x) is shown shown in the the xy-plane xy-plane above. If g(x) = (x + 3)(x ‐ 1), for above. If g(x) = + 3)(x - l ), which which of the the following ( x ) > g(x) following values values of x is is ff(x) g(x)?? A A)) ‐- 3

A A)) ‐- 5

B)) -‐ 2 B

B B)) -‐ 1

C) 1

C) 1

D) 2

D)) 5 D

In the xy-plane, the graph of the function g is the image of the graph of the function f after a translation of 11;~ units translation right. Which units to the the right. Which of the the

following g(x) ? following defines defines g(x) ? A) ) g(x) A g u )==J(3x f 6 r-‐ 2) E B)) gg(x) f(3x + 2) B ( x ) = fGX+fl C)) gg(x) 3) C ( x )= = ff(2x o ‐- w D) g(x) = f(2x 3) D ) g u ) = f o ++%

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THE COLLEGE PANDA THE COLLEGE PANDA

w”

y

y

A

2 The the function is shown The graph graph of the function y = = 9 -‐ xx2 shown in AB ?? the xy-plane the length the xy-plane above above.. What What is the length of E

The function f(x) = x3 + 1 is graphed in the xy-plane above. If the function g is defined by g(x) = x + k, where k is a constant, and J(x) = g(x) has 3 solutions, which of the following could be the value of k?

A) 3v'2 A) 3\/§

B ) 3v'10 3m B) C) 9

D ) 9v'10 9m D)

A)) -‐ 1 A B) 0 C) 1

D) 2

y

[n the -+- 12, 12, where where 1n the xy-plane, xy-plane, the the function function y = ax ax + constant, passes through the the point point ((‐a,a). - a, a) . a is a constant, passes through If a > value of a ?? > 0, what what is the the value

The function f is graphed in the xy-plane above. If the function g is defined by g(x) = f(x) + 4, what is the x-intercept of g(x)?

A) - 3 B) - 1 C) 3

D) 4

145

17

Quadratics Quadratics Just as lines were one group of functions that that have o w n properties, quadratics are are another. another. A A quadratic quadratic Just as lines were one group of functions have their their own properties, quadratics is a function in the is a function the form form f(x) = r - ax axz2 + bx b x+ J(x) +c in which which the highest power power of x is is 2. 2. The The graph graph of of aa quadratic quadratic is the highest is a a parabola. parabola .

To review we’ll walk walk through through aa few examples various properties properties you you need need to to review quadratics, quadratics, we'll examples to to demonstrate demonstrate the the various know . know.

QUADRATIC 1: QUADRATIC 2 -- 44x f(x) = xx2 x ‐- 21 f(x) 21

The Roots The Roots The The roots roots refer refer to the the values values of x that make Jf ((x) x) = : 0. They're They’re also x‐intercepts and and so solutions. We’ll that make also called called x-intercepts lutions . We'll mainly use the mainly use the term term "root" “ r o o t ” in this this chapter, the other other term termss are just as Don’t forget that that they they all all chapter, but but the are just as common. common. Don 't forget mean the the same mean thing.. Here Here,, we we can just factor find the roots: same thing can just factor to to find the roots: 2 - 4x :00 xx2~4x‐21 - 21 =

( . r-‐ 77)(x ) ( . r+ + 3) 3 ) =2 0 (x x X

: 7, = 7, -~33

The roots roots are are 7 and The this means means the the quadratic quadratic crosses the x-axis x-axis at at xx =z 77 and and x.r =: -A3. and -43. 3. Graphically, Graphically, this crosses the 3.

The Sum Sum and Product of the Roots Roots

=

We already found We already found the so their u m is just just 77 + ( -‐ 33)) z 4 their product product is is just just 77 xx -‐33 = z -‐ 221. 1 . This This the roots, roots, so their ssum 4 and and their was really easy, so was really easy, about these these values? values? Because Because sometimes sometimes you'll you’ll have have to find the the sum s u m or or the the so why why do we care care about product the roots roots without without knowing knowing the the roots roots themselves. themselves. H o w do we do that? product of the How do we do that? . . . b Given quadratic of the form y = : ax axz2 + + bx + c, the the sum s u m of of the of the the roots roots Given a a quadratic the form the roots roots IS is equal equal to to -‐ ~E and and the the product product of a

, c . equal c. is 1s equa 1to -E.

a

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THE COLLEGE COLLEGE PANDA THE PANDA

our example, example, a = In our : l,Lbb = : -‐ 44,, c = -‐21.So, 21. So,

Sumz‐é=‐‐:4 = - -- 4 = 4 a 1 l

b Sum = - -

C Product Product = -E = --‐ ‐-_21fl = ‐- 2211

1 l

a

See how were able these values values without without knowing knowing the roots that that we we how we were able to determine determine these the roots roots themselves? themselves? The The roots found found earlier earlier just just confirm confirm our our values. values.

The Vertex The

The vertex The vertex is the the midpoint midpoint of a a parabola. parabola .

y

vertex vertex

The x‐coordinate The vertex is always always the midpoint of the ttwo w o roots, roots, which found by averaging them. x-coordinate of the the vertex the midpoint which can can be be found by averaging them. Because roots are are 7 7and the vertex vertex is at x = z Because the the roots and -‐ 33,, the

7+ + (-( ‐ 33))

z 2. When When x = : 2,f(x) (2)22 -‐ 4(2) 5. = 2, J (x) := (2) 4(2) ‐- 21 21 := ‐- 225. 2 Therefore, the Therefore, vertex is at (2, -‐25). Note that that the maximum or m i n i m u m of a quadratic always at at the vertex. the vertex 25). Note the maximum minimum quadratic is is always the vertex. ln this this case, in minimum of ‐25. case, it’s it's a minimum - 25.

Form Vertex Form Just asslope-intercept as slope-intercept form Just b)is of representing form is is one one way way of of representing representing form (y = mx+ mx + b) is one one way way of representing a a line, line, vertex vertex form a quadratic quadratic function a function.. We've We’ve already t w o different different ways ways quadratics can be be represented, namely standard already seen seen two quadratics can represented, namely standard form (1; (y = form : ax2 form (y = : (x -‐ a a)(x looks likey = a(x a(x -‐ h)2 + k. ax2 + + bx bx++ c) and and factored factored form )(x -‐ b)). Vertex form form looks like y = h)2 + To get get a a quadratic quadratic function function into into vertex vertex form form,, we we have have to to do do something something called called completing completing the the square. square. Let’s Let's walk walk through through it step-by‐step: step -by-step: yy z= x2 x2 -‐ 4x 4x -‐ 21 21 the middle See the The -‐ 44.. That That’s the key. key. The The first step is to divide divide it by 2 Then write the middle term? term? The 's the step is it by 2 to to get get -‐ 22.. Then write the

following: following :

y = (x ‐- 2)2 2) 2 ‐- 21

where we put See where ? The The first part o w the take that that -‐22 and square it. We get get put the the ‐- 22? part is done. done . N Now the second second step step is to take and square it. We 4. yy:= (x‐2)2‐21‐4 (x - 2)2 - 21 - 4 See where where we Wesubtracted at the end. The The vertex we put put the the 4? 4? We subtracted it at the end. vertex form form is is then then

y z= (x -‐ 2)2 -‐ 25

To recap, recap, divide middle coefficient to get get the the number divide the middle coefficient by by 22 to number inside inside the the parentheses. parentheses. Subtract Subtract the the square square of of that at the the end. that number number at end .

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CHAPTER 17 QUADRATICS CHAPTER 17 QUADRATICS

Completing the the square square takes Completing takes some and practice, practice, so so if if you some time time and you didn't didn 't catch catch all of of this, this, first first prove prove to to yourself yourself that it is indeed that the same same quadratic the result. result. Then indeed the quadratic by by expanding expanding the Then repeat repeat the the process process of of completing completing the the square square yourself. If you’ve yourself. slightly different different way, way, feel free to many m o r e examples in this this you've been been taught taught a a slightly to use use it. We’ll We'll do do many more examples in chapter. chapter. N o w why vertex form? look at at the the numbers! It’s called vertex form reason. The The Now why do do we we care care about about vertex form? Well, look numbers! It's called vertex form for for aa reason. vertex (2, -‐25) vertex be found found just just by looking at the the numbers numbers in found the the vertex, 25) can can be looking at in the the equation. equation. But we we already already found vertex, you say! Yes, that’s true, but had to find the the roots roots to to do do so finding the o t always you that 's true, but we had so earlier, earlier, and and finding the roots roots is is nnot always so so easy. Vertex form vertex without knowing the quadratic. It’s also very much form allows allows us us to to find the the vertex without knowing the roots roots of of a a quadratic. It's also very much tested on the tested the SAT! SAT!

One o t e ‐ o n e of o f the the most common mistakes students make = ((xx -‐ 2) 2)22 -‐ 2 and think One final n note-one most common mistakes students make iiss ttoo look look a att y = 255 and think the the vertex is at (‐ 2, 2, -25) ‐ 2 5 ) instead instead of of (2, (2 -‐ 225). 5 ) One One pattern pattern of of thinking avoid this What vertex is at (thinking II use use to to avoid this mistake mistake is is to to ask, ask, What value of x would would make expression inside parentheses eq11al equal to zero? 0. value make the the expression inside the parentheses zero? Well, x: x =2 2 would would make make xx ‐- 2 2 equal equal to to 0. Therefore, vertex is at xx‐ = ‐ 2. T h i sis thinking you get the the solutions solutions from from the Therefore,the the vertex This is the the same same type type of of thinking you would would use use to to get the factored factored form form y‐‐ y = (x ‐- a)(x a)(x -‐ b). b).

The Discriminant The Discriminant

If then the As we l l explain explain later, the If a a quadratic quadratic is in the the form form ax2 ax2 + bx + c, c, then the discriminant discriminant is is equal equal to to b2‐ b2 - 4ac. 4ac. As we '’ ll later, the discriminant' Before we we explain its significance, significance, let’ the discriminant IisSa a component component of of the the quadratic quadratic formula formula. Before explain its let'ss calculate calculate the discriminant for oour discriminant u r first example, example, 2 f(x) f(x) = xx2‐4x‐21 - 4x - 21 Discriminant Discriminant = = b2‐ 4ac = : ((‐4)2 : 100 b2 - 4ac - 4 )2 -‐ 4(1)(‐21) 4(1) ( - 21) = Now, what does the the discriminant di scriminant mean? Now, what does value of the doess not n o t matter. mean? Well, the the value the discriminant discriminant doe matter. What What matters matters is is sign of the the discriminant‐whether discriminant -w hether it's the sign it’s positive, positive, negative, negative, or zero. care that zero. In In other other words, words, we we don’t don't care that it’s it's 100, we we just just care care that that it's it’s positive positive.. Letting be short short for discriminant, discriminant, Letting D be W h e nDD = z O0,, When y

W h e nDD > 00,, When

y

W h e nDD < O When 0,,

y

-----0-+----there are are ttwo there w o real t w o solutions). solutions) real roots roots ((two

tthere h e r eis one one real real root. root.

X

there no real real roots. roots. there are are no

The The Quadratic Quadratic Formula Formula

we've seen, As we’ve seen, the important aspect aspect of a the roots roots are are the the most most important a quadratic. quadratic . Once Once you you have have the the roots, roots, things things like like ve rtex form form and vertex and the discriminant are are n o t as ashelpful. or work the discriminant not helpful. Unfortunately, Unfortunately, the roots roots aren’t aren't always always easy easy to to find find or work with. That’s That's when when vertex with. u m / product of the can get get us to the the answer vertex form, form, the the discriminant, discriminant, and and the the ssum/product the roots roots can us to answer faster. But if if we we m u s t find the there is always always one one surefire surefire way way to o ‐ t h e quadratic must the roots, roots , there to do do sso-the quadratic formula. formula.

_ ‐b:|:\/b2‐4ac - b± Jb 2 - 4ac

x=

_T

2a

forax2+bx+c:0. for ax2 + bx + c = 0.

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THE COLLEGE COLLEGE PANDA THE PANDA

For the purpose quadratic formula purpose of learning, learning, let’s let's apply apply the the quadratic formula to to our our example, example,

f(x) = x 22‐- 44xx ‐- 2211 J(x) According to the According the formula, are formula, the the roots roots //ssolutions olutions are

_4 m _= 44 x± 1100 _= ?or _ x_ = -‐(‐4)i,/(‐4)?‐4(1)(‐21) (- 4) ± J( - 4)2 - 4(1)( - 21) = 4 ± i/100 x

‐ T ‐ 2(1)T ‐ ‐ ‐ 27 0 2 r ‐ 33

These values that that we factoring.. These are are the the same same values we got got through through factoring

Notice 4116, is is tucked tucked under under the o w does does this Notice that that the the discriminant, discriminant, 172 b2 -~ 4ac, the square square root root in in the the quadratic quadratic formula. formula. H How this help us know about the discriminant? help us understand understand what what we we know about the discriminant?

takes effect effect and and we we end w o different Well, when when b2‐ b2 - 4ac > 0, 0, the the"“ :±l :"" takes end up up with with ttwo different roots. roots. When When 172 b2 ‐- 4ac 4ac = = 0, 0, the the ”" ±i "” does does not n o t have we’re essentially essentially adding of which which give give us us the same have an an effect effect since since we're adding and and subtracting subtracting 0, 0, both both of the same root.. When square root root of gives us us root When [72 b2 ‐- 4116 4ac < O, 0, we’re we're taking taking the the square of a a negative negative number, number, which which is is undefined undefined and and gives no real no real roots (we’ll talk number in in aa later roots (we'll talk about about imaginary imaginary number later chapter). chapter).

Hopefully, you understand understand where meanings come come Hopefully, the the quadratic quadratic formula formula helps helps you where the the discriminant discriminant and and its its various various meanings from. Understanding from. will help you remember remember the Understanding this this connection connection will help you the concepts. concepts. Now that Now we’ve taken you on tour through that we've taken you on aa thorough thorough tour through the the properties properties of of quadratics, quadratics, we’ll we'll go go through through a a few few more examples more pace. examples to illustrate illustrate some some important important variations, variations, but but we’ll we'll do do so so at at a a much much faster faster pace.

QUADRATIC 2: QUADRATIC 2:

ff(x) (x) = : -x ‐x22 ++ 66xx‐- 1100 The Roots Roots This quadratic n d in fact, fact, if if we we look look at the discriminant, quadratic cannot cannot be be factored. factored. A And at the discriminant,

ac = : (6) (6)22 -- 44(-1)( ( ‐ 1 ) ( ‐-110) 0 ) := ‐-44 b1922 -‐ 44ac

it's negative, negative, which it’s are no no real real roots roots or solutions. solutions. The The graph the quadratic which means means there there are graph of of the quadratic makes makes this this even even more more clear:

y

2 term When the coefficient coefficient of the When x2 negative, the the parabola shape of U.” the x term is negative, parabola is is in in the the shape of an an upside-down upside-down ”"U."

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CHAPTER 17 CHAPTER 17 QUADRATICS QUADRATICS

The The Sum Sum and and Product Product of the Roots Roots

x ) = -‐x2 10 Jf ((x x 2 +66xx -‐ 10 b 6 Sum = - - = - - = 6 a -‐11 a C - 10 Product P r o d u c=t : -5 = z -_‐10 :=110 0 a -‐11

Sum= ‐9 = ‐i =6

what!? We already already determined determined that that there there were were no no roots roots.. How H o w can sum and and aa product Wait, what!? can there there be be a a sum product of roots roots that don’t exist? exist? Well, the the quadratic quadratic doesn't doesn’t have have any any real roots, have ima imaginary roots. The values roots, but but it does does have ginary roots. The values that don't above are are the the sum sum and and product product of these these imaginary imaginary roots roots.. We'll We’ll cover numbers in a above cover imaginary imaginary numbers a later later chapter. chapter.

Vertex Form Because the roots roots are are imaginary, imaginary, we we can't can't use use their their midpoint midpoint to find these cases, cases, we we must must get get the the Because find the the vertex. vertex . In these quadratic quadratic in vertex vertex form. form. We'll We’ll have have to complete complete the the square square..

yyz‐x2+6x‐10 = - x 2 + 6x - 10 First, multiply multiply everything everything by negative negative 11 to get get the the negative negative out out of the Having the the negative negative there makes First, the x2 x 2 term. term . Having there makes later . things needlessly complicated ll multiply back by -‐11 later. things needlessly complicated.. We' We’ll multiply everything everything back

-‐y:x2‐6x+10 y = x 2 - 6x + 10 get 9. Remember put the the -‐33 inside the Divide term by 2 to get -‐33 and Divide the the middle middle term and square square this this result result to get Remember that that we put inside the parentheses pieces in place, place, parentheses with with x and and subtract subtract the 99 at at the the end. end. Putting Putting these these pieces = (x‐3)2+10‐9 -‐ y =( x - 3) 2 + 10 - 9

= (xx -‐ 3) 3)22 + +11 -‐yy =( N o w multiply everything by by -‐11 again, again, Now multiply everything

: -‐ ((xx -‐ 3) 3)22 -‐ 1 y=

Now the vertex the graph graph is an upside-down ”U,” "U," -‐11 is the the Now it's it’s easy easy to see that that the vertex is at at (3, (3, -‐ 11)).. And And because because the an upside-down maximum value of Jf(x). (x). maximum value QUADRATIC QUADRATIC 3: 3:

2 f ( x ) = 2x 2x2+5x‐3 f(x) + 5x - 3

The Roots Roots The

Wecan can factor this quadratic to get get We this quadratic 2x22 + + 5x -‐ 3 = = 0 2x

((2x 2x -‐ 1l ) ( Xx + 33)) = O 0 x X

z 0.5, -‐33 =

The roots roots are are 0.5 and you don't don’t know know how how we we factored factored this, teaching factoring the The and -‐ 33.. If you this, unfortunately unfortunately teaching factoring from from the ground up is not n o t within the scope scope of this this book. book. Don't Don’t be be afraid afraid to look look up factoring lessons lessons and and drills ground up within the up factoring drills online online and in your your textbooks. textbooks. It's It’s an an essential essential skill skill to have have.. Just Just know o u t there there involves involves a and know that that every every method method out a little little trial and error. And if you’re ever stuck, the quadratic formula is always an option. trial and error. And you're ever stuck , the quadratic formula always an option .

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THE COLLEGE COLLEGE PANDA PANDA THE

The The Sum Sum and and Product Product of the the Roots Roots

5 b Sum = -‐B = -‐§ = : -‐2.5 Sum 2.5

a a

2 2

C -3 Product = E = _‐3 = -‐ l1.5 .5 Product 22 a

a=

The The Vertex

Averaging the the two t w o roots roots to find find the x-coordinate of the the vertex, Averaging the x-coordinate vertex, 0 5+ + (( ‐ 33)) =_ -‐2.5 0.5 2.5 =_ _ 25 _ 2 22 2 _ 1.25 1. Plugging Plugging this this into into f(x) f (x) to find find they-coordinate, the y-coordinate,

2(‐1.25)2 : ff(‐1.25) (- 1.25) = 2( - 1.25)2 + 5(‐1.25) 5( - 1.25) -‐ 3 3 =

-‐6.125 6.125

The vertex vertex is at at (( ‐ l 1.25, . 2 5 , -‐6.125). Because the the quadratic quadratic opens shape of aa "U," ” U , ” the the minimum The 6.125). Because opens upward upward in the the shape minimum value is -‐6.l25. 6.125. value of J(x) f ( x ) is

Vertex Form : 2x 2x22 + + 5x -‐ 3 y= 2 First, square, always make sure sure the the coefficient coefficient of xx2 First, divide divide everything everything by 2. Before Before completing completing the the square, always make is l. We'll We'll multiply multiply the the 22 back back later later.. y 5 3 2

2=X

+ 2X- 2

25 Divide result to get get~!. We put put the the~2 inside the parentheses with Divide the the middle middle term term by 2 to get~ get 2 and and square square this this result 1‐6. We inside the parentheses with

25 16

x and and subtract subtract the the 3‐2 at at the the end. end.

-"‐ We:re deal~g with the intersection of two gra hs a ain. So wh of consisting Of the system solve the We solwg we do . We at do th fj ti~ . g the eguat10ns. their the" equations, Substih1ting Substituting the the second second eq equation system C°"5i5ti"8 get we get rst, we into e first ua on into 1 0=: -‐x2+6.r+3 x + 6x + 3 10 2

+ 6x 6x‐- 7 7 x2 + = -‐x2 00z Now and finish this to intersection point(s) just like previous the previous in the did in we did like we point(s) just the intersection find the to find solving this finish solving ~,ead and go ahead couldgo we could Now we example, but there’s a faster way. For the purposes of this question, we don ’t care where the intersection points points intersection where care 't don we question, this of purposes the example, but theres a faster way. For are. We just want to know how many there are. there many how know want just We are. that. the discriminant Sound discriminant to do do that. use the can use We can familiar? We Sound familiar? 2 4(: (6) (6)22 ‐- 4 ( ‐ 11)() ( ‐ 77)) := 8 Discriminant: b2‐ - 4ac = Discriminant = b

there are 2 above . If there up above. we set up equation we are 2 solutions there are The discriminant which means solutions to the equation means there positive, which discriminant is positive, bother finding didn't bother points . To summarize, there must solutions must be be [I} intersection intersection points. summarize, we didn’t finding above, there equation above, the equation solutions to the might've points intersection the and care, we all for 100 = x and 2 = been x could've been the two and and the intersection points might’ve values of x. They could’ve two values the used the we used and we them, and of them, were ttwo there were that there mattered was doesn't matter. ). It doesn’t been and (100,6 (100,6). matter. What What mattered was that w o of (2,5 ) and been (2,5) And point. intersection one be only would that. If the discriminant discriminant determine that. discriminant were were 0, there there would be one intersection point. And ifif discriminant to determine . points intersection points. no intersection be no would be there would than 0, there were less than discriminant were the discriminant points intersection points the intersection where the out where and figure out back and understand this question you understand Make sure Make sure you question.. Feel free to go go back so was so discriminant was the discriminant why the That 's why formula . That’s need the fun . You'll not fun. It's not (Hint: It's are (Hint: actually are actually You’ll need the quadratic quadratic formula. helpful). helpful).

EXAMPLES: EXAMPLE3:

y -‐ kk = = 0O

= x2-3x + 1 yy=x2‐3x+l system the system does the k does of It the following of the For which constant. For a constant. is a k is above, k equations above, system of equations the system Jn the in which of following values values of solutions? equations have of equations have no no real real solutions? A) ‐22 A)-

B) -‐11

C)O

D) 1

equation , second equation, the second into the equation and we get y = First, we First, z k from the first equation and substitute substitute this this into

kk = x2 +1l 3x + x 2 -‐ 3x O=z x22-‐ 33xx + ((11-‐ kk)) O The real solution no real have no should have above should equation above the equation then the solution, then real solution, no real has no equations has system of equations If If the system solution.. The than 0. be less than discriminant should discriminant should be 4k 4k = 4(1)(1 -‐ k) = 9 ‐- 4 + 3)2 -‐4(1)(1 b2 -‐ 44ac Discriminant = 172 Discriminant: a c = ((‐3)2 +4k z 55 + +4k

answer 2, answer Only -~2, negative. Only being negative. 4k being + 4k 5+ in 5 which one to see choices to each of the test each Now N o w we test the answer answer choices see which one results results in discriminant. negative discriminant. a negative produces a ~ produces back Go back SAT. Go the SAT. on the see on might see you might questions you toughest questions the toughest far showcase so far done so 've done examples we The examples we've showcase some some of of the them . understand them. make sure and make and sure you you understand 154

THE COLLEGE PANDA THE COLLEGE PANDA

EXAMPLE n) = ‐100n2 - 10Qn2 + 000n to model model the population EXAMPLE 4: A biologist biologist uses uses the the function function p( p(n) + 1, 1,000n population of seagµlls seagulls

equivalent forms of p(n) p(n) displays on aa beach beach in year number number n, where where 1 ~ 5 n ::; S 10. Which of the the following equivalent displays which the population reaches that the the maximum maximum population population of seagulls seagulls and and the the number number of the year in which population reaches that maximum maximum as as constants constants or coefficients? ‘ A) 4n(25n -‐ 250) A) p(n) = -‐4n(25n

B) 10(10n2 -‐ 100n) B) p(n) p(n) = -‐10(10n2 10011)

C) p(n) = -‐100(n lOO(n-‐ 5)2 + 2,500 C)

D) lOO(n- 7) 4,900 D) p(n) = -‐100(n 7)22 + +4,900 Anytime maximum or minimum output (i.e. (i.e. Anytime you you see see a quadratics quadratics question question that deals deals with with the maximum minimum of a function function output the y-value), either figure out the vertex or look for vertex form. After all, the the maximum y-value), either figure look After all, the vertex vertex is where where the maximum or minimum minimum occurs occurs.. In fact, fact, the the answer answer is is either either (C) (C) or (D) because those are are the only only ones vertex form form.. because those ones in vertex Furthermore, with that (D) does does not the original original equation Furthermore, with a little little calculation, calculation, it's it's easy easy to see that n o t expand expand to be be the equation,, so so the the answer answer is (C).

However, learning purpo ses (and questions), I'll this question question in ttwo However, for learning purposes (and for the the tougher tougher questions), I’ll show show you you how how to do do this wo different ways ways.. We We can can find the the vertex vertex using using the the average average of the roots roots and vertex form form.. different and then then reverse reverse engineer engineer vertex Or we transform the the equation we can can transform equation into into vertex vertex form form directly. directly. Solution and factor, Solution 1: To find the roots, roots, we we set set the the equation equation equal equal to O 0 and

0 -400,12 100n 2 + 1,000n 1,000n = 0 -‐100n(n lO0n(n -‐ 10) = z 0 : 0,10 n= 0, 10

The roots vertex is 5. Now roots are are O 0 and and 10, which which means means the the x-coordinate x-coordinate of the vertex Now we we can can plug plug 5 into into p(n) to find find the y-coordinate.. they-coordinate : -‐100(5)2 +1,000(5) : 2,500 2,500 p(5) = 100(5) 2 + 1, 000(5) = So 2500). Now So the the vertex vertex is at (5, (5,2500). N o w remember remember what what vertex vertex form form looks looks like: y = : a(x -‐ h) h)22 + + k. Given Given our o u r values, values, we have have = a(n -‐ 5) 5)22 + + 2,500 p(n) = 2,500 We now another point point to work work with easy to see that n o w need need to find what what a is. To do that, that, we we need need another with.. Well, it's it’s easy that p(n) passes ). Plugging passes through through the the point point (0,0 (0, O). Plugging that that in, in, 0= z a(O a(0 -‐ 5)2 + + 2,500 2,500 o 2 25a 2511 + + 2,500 0= 2,500 2,500 -‐25a 25a = 2,500

a == -‐100 100

[@[J .

Finally, p(n) = -‐100(n 5)22 + 2, 2,500. Answer (C) . Finally, lO0(n - 5) 500. Answer

155

QUADRATICS 17 QUADRATICS CHAPTER CHAPTER 17

divide First, divide directly . First, form directly. vertex form the square involves completing method involves second method This second Solution completing the square to get the vertex Solution 2: This l. coefficient of 1n122 is 1. the coefficient ensure the - 100 to ensure everything by ‐100 everything

p(n)) = 1,00011 + 1,00011 - 100n2 + = ‐100n2 p(n POI) _ 2 - lOn p(n) = ‐‐_100 ‐n n2_ lOn - 100 lOn. be -‐10n. would be term would "middle" term end, the ”middle" what to do next? you remember Do you remember what next? If If we wrote wrote the the constant constant 0 at the end, and the with n and parenthe ses with the parentheses inside the belong s inside - 5 belongs that to get and square get -‐55 and the ‐- 110 Divide the Divide 0 by 2 to get square that get 25. The The ‐5 end . the end. at the subtracted at 25 gets subtracted 25

p01) _ (n 25 (n -‐ 5)2 p(n) = W _ 5)2 ‐- 25 - 100 - 100. back by ‐100. everything back multiply everything can multiply we can Now N o w we 2,500 + 2,500 p(n) = ‐100(n 5)22 + - l00 (n -‐ 5)

. [@IJ

that the again, we prove And again, And prove that the answer answer is (C) .

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Review: Review:

+ c,c, ax2 + bx form, y = the form, Given = ax2 bx+ quadratic of the Given aaquadratic ways: following ways: the following be found The roots, also also called called solutions solutions and and x-intercepts, x‐intercepts, can can be found in the The roots,

0 Factoring • Factoring

0 on the the calculator calculator (look for the the x-intercepts) x‐intercepts) (look for Graph on • Graph

o The quadratic formula x = quadratic formula • The

m -‐ bb± ✓bi2 - 4ac 2a 211

= - -ab

Sum of the the Roots Roots = ‐5 Sum

C

Product of the Roots = = -2 the Roots Product a

2 - 4ac ThediscriminantD =b b2‐4ac The discriminant D =

solutions. 0 When D > > 0, w o real real solutions. are ttwo there are 0, there • When one real When D = 0, there is one real solution. solution. 0, there •0 When solutions. real solutions. 0 are no no real there are 0, there When D < 0, • When the vertex, vertex, find the To find get to get quadratic to the quadratic that value Then plug 0 average of the the roots roots to get get the the x-coordinate. x-coordinate. Then plug that value into into the the average Take the • Take the y-coordinate. y-coordinate. the square . the quadratic Put the quadratic in vertex vertex form form by by completing completing the square. •0 Put 2 is positive 1 by dividing everything by a. coefficient of xx2 Ensure the the coefficient positive 1by dividing everything by 1. Ensure

C b 2 y =x2+ x+ +x -a = X + -E a a

b b2 b that result Square that 2a. Square get 2. term bx to getZa middle term the middle 2. Divide Divide the the coefficient coefficient of the result to to get g e4a1,22 t 2‐ .. PPutu 2a 2a t ‐ 2.

!

2

end. the end. subtract £72 at the with x and inside parentheses with and subtract the parentheses inside the

z_(.+£)2+£_i +:. - ~ (x+ _!:_)2 a 4a 2a a

¥_= a_

211

by a. 3. everythingby Multiply everything 3. Multiply

2 b 2

a

4:122

2 b2

c ‐ ‐b ‑ = a ( x +-b ) + c-yy=a(x+‐‐2) 4a 2a

actual with actual quadratics with on quadratics variables. Practice the variables. with the 4. IIt's t ’s unnecessary unnecessary to memorize memorize these these steps steps with Practice on k. + )2 h a(x = y like: looks form vertex numbers. However, do do remember remember what what vertex form looks like: y : a(x‐- h)2 + k. numbers. However, a quadratic, find minimum or the minimum asked for the you're asked Whenever or the the maximum maximum of aquadratic, find the the vertex. vertex. Whenever you're

157

QUADRATICS CHAPTER 17 QUADRATICS CHAPTER 17

CHAPTER EXERCISE: Answers for this chapter start on page 314.

the used on the be used NOT should N calculator should A calculator O T be following questions. following questions.

3x2+10x=8 3x 2 + lOx = 8 equation the equation solutions to the If w o solutions the ttwo are the and b are If a and ? b2 ? value of If2 the value above what is the > b, what and a > above and

the between the the distance what is the xy-plane, what the xy-plane, In the distance between parabola the parabola x-intercepts of the ttwo w o x-intercepts yy = x2 10?? 3x ‐- 10 x2 ‐- 3x

4

A) 9 2 B) 3

A) 3 B) 5

C) 4 C) 16 D) 16 D)

C) 7

D)) 110 D 0

fJ(x) (x) = =m[(x‐m)2‐l] m [(x - m) 2 - 1]

to x 2 + 4x + 2 = 0 ? solutions tox2+4x+2=0? the solutions What are the Whatare

positive a positive m is a above, m defined above, hmction ff defined the function In the

A i fi = z- 2‐ ±Z vf2 A) ) x x B n/i B)) x x= 2z±e2v12 C) ) Xx=C = ‐ 22 i±22y12 \/§ D = ‐- 44 i±Z2,/2 \/§ D)) xx=

xy-plane is a the xy-plane offf in the graph of The graph constant. The constant. following statements the following parabola. statements Whkh of the parabola . Which true? parabola is true? the parabola about the about ( m, ‐- 11). at (m, occurs at minimum occurs A) [Its t s minimum ).

- m). at (m, ‐m). occurs at minimum occurs Its minimum B) Its B) 1). (nz, ‐- 1). at (m, occurs at C) maximum occurs Its maximum C) Its

m). (m, -‐ m at (m, occurs at maximum occurs Its maximum D) Its D) ).

= 0, what is the value 7a + 3 =O,whatisthevalue 2a 2 ‐- 7a+3 lIff a < 1 and 2a2 l and o a?? off a yy = ‐- 33 2 yzx2+cx y = x + ex

constant. a constant. c is a above, 0is equations above, system of equations In the system ln the the does the values of c does following values the following For which of the For which real exactly ttwo have exactly equations have system w o real system of equations

solutions? solutions? solutions of the solutions the ssum What u m of the What is the 2 =4x+5? (2x‐3)2 (2x - 3) = 4x + 5?

A A)) -‐ 4 B) 1

C) 2 D) 3

158

THE PANDA THE COLLEGE COLLEGE PANDA

A calculator is allowed on the following questions.

m 2 ‐- 100m -‐ 120,000 P= = m2 120,000

The The monthly monthly profit profit of a mattress mattress company company can can be modeled modeled by the equation equation above, above, where Pisis be by the where P the profit, profit, in dollars, dollars, and and m is the number of the the number mattresses sold. the minimum minimum number number mattresses sold. What What is the of mattresses sell in a given mattresses the the company company must must sell a given month so that it does not lose money during that month so that does not lose money during that month? month?

At which the following following points does the which of the points does the line line the parabola parabola with with equation equation y = z 4 intersect intersect the y = (x + 2)2 2)2 -‐ 55 in the xy-plane? the xy-plane? A) 1,4 ) and A) ((‐1,4) and (( ‐ 55,, 4 ) B) 5,4 ) B) (1,4 (1,4)) and and ((‐5,4)

C) (1,4 ) and 4) C) (1,4) and (5, (5,4) D) 11, 4) and D) ((‐11,4) and (7,4 (7,4))

y

E(x) = 50x 50x22 ‐- 8003: + 10,000 10,000 E(x) = B00x + The function function above above models the relationship relationship The models the between the the total total monthly expenses E, in dollars, dollars, between monthly expenses its restaurant and and the the number tables x in its of a restaurant number of tables dining area, area, where where O::; What does does the the dining O5 x::; x 5 25. What number 10,000 represent the function? function? number represent in the (3, -8)

maximum number tables that A) The The maximum number of tables that can can fit in the the dining dining area area

Which of the following equation s represents the parabola shown in the xy-plane above? A) y

= (x - 3)2 -

8) y =(x+

C) y

= 2(x -

average monthly expenses, in dollars, dollars, B) The The average monthly expenses, for each table in the dining room room each table the dining C) The dollars,, The total total monthly monthly expenses, expenses, in dollars when there there are tables in the dining when are zero zero tables the dining

8

3)2+ 8

area area

3)2 - 8

D) The The total total monthly monthl y expenses, expenses , in dollars, dollars, when the the number number of tables tables in the the dining dining when area maximized area is maximized

D) y= 2(x+ 3)2- 8

2 For value oft of t does the equation equation vv = St St -‐ tt2 does the For what what value result in the maximum value value of v ? result the maximum

f (x) = - x 2 + 6x + 20 The defined above above.. Which the The function function f is defined Which of the following forms of f (x) displays the (x ) displays the following equivalent equivalent forms maximum value of f as maximum value as a a constant constant or coefficient? coefficient?

A ( x )=: ‐-(( xx‐-3 3)2 ) 2 ++ 11 A)) ff (x) 11 2 B ) f ( x ) : ‐ ( x ‐ 3 ) 2 + 2 9 B) f (x) = -(x - 3) + 29 C) f(x)=‐(x+3)2+11 C) f(x) = - (x + 3)2 + 11 D) D) ff(x):‐(x+3)2+z9 (x)=-(x+ 3)2+ 29

159

CHAPTER 17 QUADRATICS CHAPTER 17 QUADRATICS

yy = = ‐- 33

g(x)

= - 3x 2 + 18x

The gives the the data The function function 3 g above above gives data transfer transfer speed, second, over network speed, in megabytes megabytes per per second, over a a network connection connection x minutes minutes after after a file transfer transfer was was initiated. g(x) in the xy-plane initiated. The The graph graph of y = = g(x) the xy-plane has c. Which Which of of the has x-intercepts x-intercepts at at x = 0 0 and and x = c. the following is the following the value value the best best interpretation interpretation of the

y=ax2+4x‐4 y = ax2 + 4x - 4 In the system of equations constant. the system equations above, above, a a is aa constant. For which of the the following For which does the following values values of a a does the system of equations equations have system have exactly exactly one one real real solution? solution?

cC

_7 ?. of 2

A)) -‐ 4 A

2

B)) -‐ 2 B

A) The The initial initial data data transfer transfer speed speed over over the the network connection connection network

C) 2

D)) 4 D

B) B) The speed over over the the The maximum maximum data data transfer transfer speed

network connection connection network C) The data transfer The time time at which which the the data transfer speed speed over the the network network connection over connection was was at its highest highest

f ( x ) = x 2‐ 24x +180

time at which D) The The time which the the file transfer transfer completed completed

For a a manufacturer For x‐ray machines machines,, the manufacturer of x-ray the cost per unit, in thousands per unit, thousands of dollars, can be dollars , can be modeled modeled the function function f above, above, where by the where x is the the weekly weekly number of units units produced. produced. How number How many many units units should the the manufacturer manufacturer produce produce each should week to each week minimize the the cost minimize cost per per unit? unit?

y

= a(x -

3)(x - k)

the quadratic In the are quadratic equation equation above, above, a a and and k are constants. If constants. graph of the If the the graph the equation equation in the the xy-plane is a xy-plane (5, -‐32), a parabola parabola with with vertex vertex (5, 32), what what

is the the value value of a ? ? A) 2 B) 5 C) 6

D) 8

f(x) = : -‐4x2 x J(x) 4x 2 + 2 22x The function The function ff above above gives gives the the data data transfer transfer speed, in megabytes per per second, speed, megabytes second, over over a a network network connection x minutes connection transfer was was minutes after after a a file transfer initiated. graph of y = J(x) f(x) in the xy-plane initiated . The The graph the xy-plane has x-intercepts at = 0 and = b. Which has x-intercepts at x = and x = Which of of the the following is the following the best best interpretation interpretation of b b??

In the line y = = 2x + + b intersects the xy-plane, xy-plane , the the line intersects the the parabola at the point (3,k). If parabola y = = x2 x 2 + bx + 5 5 at the point (3, k). If b b is a constant, is the constant, what what the value value of k ? A) 0

A) The the The initial initial data data transfer transfer speed speed over over the network connection network connection speed over over the the B) The The maximum maximum data data transfer transfer speed network network connection connection C) The The time time at at which which the the data data transfer transfer speed speed over a s at its its over the the network network connection connection w was highest highest The time at which D) The time at which the the file transfer transfer completed completed

B) 1 C) 2

D) 3

160

18 e

•

Division Synthetic Division Synthetic way you same way the same another in the involves dividing Synthetic dividing one one polynomial polynomial by another you divided divided numbers numbers in in 3rd 3rd division involves Synthetic division grade. . grade

xx 2 2++ 3x 3 x 7 22

1 8 R 2

3 5 6 3156

R R ‐- 1]

+ 2x2 5x + 1 xv1|Fx3+2x2~5x+1

x - 1 Ix 3

several shortcuts towards several you towards direct you then direct way first, but "mathematical" way I'll teach l’ll teach you you the long long “mathematical” but then shortcuts that that will will get get questions way. These long way. without using on the SAT without you through through almost almost any any synthetic synthetic division division question question on using the long These questions you do, they'll rarely show up, up, and and if they do, they’ll show show up up only only once once.. rarely show synthetic division. to synthetic applies to same logic applies can see how by 33 so Let’s retrace steps of dividing dividing 56 56by so you you can how the same division. retrace the steps Let's to get and a 1 x 3 = on top once . We put First, we see that that 3 goes into into 5 once. put aa 1“1on top and z 3 below below the 5. We then then subtract subtract to get 22 First, and down. bring the 6 dOWn. and bring l 6 3156

3 26 26

below the 24 below into 26? 8 times does 3 go into times does many times N o w how how many times.. So we put put an 8 up top and and a 3 x 8 = z 24 the 26. 26. Now Subtracting, Subtracting, we get 2. l 8

3156 3

2 6 2 4 - 2-

eighteen into 2. Therefore does nnot and 3 does At this point, point, there there are are no more m o r e digit digitss to bring bring down down and o t go into Therefore,, 3 goes iinto n t o 56 56 eighteen At form: following the in written be can result This two times w i t h a remainder of t w o . result can be written following remainder with times 56 = 18~

3

3

divisor . and 3 is the divisor. quotient, 2 is the remainder where 18 18is the quotient, remainder,, and is the where 161

CHAPTER 18 18 SYNTHETIC SYNTHETIC DIVISION CHAPTER DIVISION

The process The essentially the the same. same. To show you you how synthetic division division works, process of dividing dividing a a polynomial polynomial is essentially To show how synthetic works, let’s let's 3 + 2x divide divide xx3+ 2x22 -‐ Sx 5x + ll by by x -‐ l.1. 3? xx22 times. 3. We H o w many many times We How times does does x -‐ 1 l go into into xx3? times. Why? Why? Because Because xx xx x2 x2 = = x3. x3. The The goal goal is is to to match match xx3.

3- x 2 . This don’t i n " step. step. Now, Now, (x 1) xx x2 x2. what we we put don ' t care care about about the the -‐11 during during this this ”fitting "fitting in" (x ‐- 1) x2 = xx3‐ This is is what put below below dividend. the dividend.

xx22 3 xx -‐ l1x j x3

++ 22xx2 2- ‐ 5Sxx ++ 11

xx33 ‐ x x2 2 Finally, we subtract like we Finally, we subtract do in basic number division. division. Notice Notice that subtract each we do basic number that we we must must subtract each element, element, sothe so the -~x2 x2 2 2 becomes x , yielding becomes + +x2, yielding 3x2. Unlike in long long division division with dividend 3x . Unlike with numbers, numbers, all the remaining remaining terms terms from the the dividend should be be brought brought down down for each should each step step in synthetic synthetic division. division. x2 x2 xx33

xX -‐ l11 ++ 22xx22- ‐ 5Sxx ++ 11 x3 x 3 ‐ x x2 2 3 x2 2- ‐ Sx 5 x++ 11 3x 2 ? 3x Next step. step. H How many times time s. Remember Next o w many times does does x -‐ 1 go into into 3x 3x2? 3x times. o u r goal Remember our goal at each each step step is to get the same same exponent exponent and and the the same same coefficient the term with the highest the coefficient as as the term with +3x up and highest power. power . We put put the the +3x up top top and 3x x (x -‐ 1) = 3x2 3x 2 -‐ 3x 3x on on the the bottom. bottom . 3x

x2 x 2++ 33xx

+ 22xx2 2- ‐ 5 + 11 Sx x + xx -‐ 1 l jxx33 + xx33 -‐ x x2 2 3 x2 2 ‐ Sx 5 x++ 11 3x 3 x2 2 ‐ 33xx 3x And just like last last time, subtract each each term, term, nnot And time, we subtract o t just just the 1. the first. We then then bring bring down down the the 1. x2 x 2 ++ 33xx

x ‐ 111 xx33 ++ 22xx2 2- ‐ 5 + 11 Sx x + x3 x 3 ‐ xx2 2 3 x2 2- ‐ Sx 5 x++ 11 3x 2 - 3x 3x2‐3x 3x -‐ 22xx + + 1 1

X -

162

THE COLLEGE COLLEGE PANDA THE PANDA

We’re How many many times does xx -‐ l1 go go into 1? -‐22 times times.. So up top and We're almost almost done. done. How times does into -‐ 22xx + + 1? So a a ‐2 - 2 goes goes up top and 1) = = -‐ 22xx + 2 goes the bottom. bottom. -‐22 x (x -‐ 1) goes on on the x2 x 2+ + 3x 3

x ‐ 22

x33 ++ 22xx2 2 ‐ 5 5xx ++ 1 xx -‐ 1 l lx 1 ‘% xr :1

‐

x2 x2 2 2 ‐ 5x 3xx 3 5 x++ 11 3x 2 3x2‐3x 3x

‐ 22xx ++ 1 1 ‐ 22xx ++ 2 Subtracting, we Subtracting, we get get ‐1 - 1 at at the the end. end . x2

X -

+ 3x 3 1 1x + 2x 2 x3 3 : 3 ‐ xx2 2

2 5x +

1

2 2 ‐5x 3x x 3 5 x+ + 11 3x 2 3x2‐3x 3x

‐ 22xx ++ 1 ‐ 22xx ++ 2 -‐11

. . 56 2 . . 56 know we're we’re done With a constant. constant. And 185, m1xed fraction, fraction, We know done when when we end end up with And Just just as as we can can express express ‐3‐ as as 18~, aa mixed 3 we we can can express express x33++22xx 22_- 5x 5 x+ + 1 as x2+3x‐2‐ 2 11 as X + 3x - 2 - -x -‐ 1l x -‐ 1l Notice where Notice each component component is placed. placed. The The quotient u t in remainder, -‐ 11,, is is the the where each quotient is written written oout in front. front. The The remainder, numerator of the numerator fraction and the divisor, x ‐ 1, is the denominator. These placements are exactly the same as the fraction and the divisor, - l, is the denominator. These placements are exactly the same as long division in long Get used used to to seeing seeing synthetic results in in this division with with actual actual numbers. numbers . Get synthetic division division results this format. format.

Here’s another that’s the the same. same. The The result result of of our o u r long long division numbers Here's another thing thing that's division with with numbers 1 8 R 2 3 5 6

3156 means 56 = 18 + 2. means that that 56 = 3 x 18

The same meaning applies our synthetic synthetic division The same meaning applies to to our division result. result .

x3+2x2‐5x+1=(x‐1)(x2+3x‐2)‐1 x 3 + 2x 2 - 5x + 1 = (x - l ) (x 2 + 3x - 2) - 1 Dividend = = Quotient Quotient x Divisor Remainder Dividend Divisor + Remainder Hopefully you've you've been to grasp synthetic division o r e intuitively Hopefully been able able to grasp synthetic division m more intuitively through through the the comparison comparison with with regular regular long the parts each other other in in the where long division. division. A Alll l the parts relate relate to each the same same way. way. Let's Let's dive dive into into some some more more examples examples where we show you we can can show you some some shortcuts. shortcuts.

163

CHAPTER CHAPTER 18 18 SYNTHETIC SYNTHETIC DIVISION DIVISION

6 5 EXAMPLE expression (2:25 x - is equivalent the following? following? EXAMPLE 1: The The expression equivalent to which which of the

x+ 2

17

A’6“m x+2

A)6 -~

B)6+

7 _2_

B)6+x+2 x+2

6 -‐ 5 C) 6-5

07

5 0)6 - -

9’6‘52

2

Using division, Using synthetic synthetic division, 6

x X

+ 22 I6x 6x +

‐

5

6x + + 12 6x

‐

17 17

The quotient quotient is 6 and and the the remainder remainder is -‐ l17. 7 . We We can can write write this this result xl‐ZZ'-An Answer The result as6 as 6 -‐ ~ swer~-(A) . x +2 ~ Now question without division? N o w how how would would we we approach approach this this question without using using synthetic synthetic division? 6xx-‐ 5 6(2) 77 . 6 6 ( 2 )- ‐ 5 We up.. Let's say x == 22.. ' IThen w e ccan a nplug p1u g in l n numbers n u mb ers that t hat we w e make m ak eup Let ’ssayx ' he n --x + 2 = ---2 + 2 = ‐-4 . x+2 2+ 2 4 7 n o w look look for an an answer answer choice choice that that gives gives 2 when when x = z 2. We can rule out right away away since We now can rule out (C) and and (D) right since they they

4

i·

don't give give 2. Plugging Plugging x = z 2 into into answer answer (A) gives gives don't

_17_617 x+2 _

4

2417 4 4

7 4

This making up numbers and testing each each answer This confirms confirms that that the the answer answer is indeed indeed (A) (A).. This This strategy strategy of making up numbers and testing answer choice be much choice can can be much faster faster than than synthetic synthetic division. division. 7 EXAMPLE 2: When 3x 2 + 4 is divided by x - 1, the result is A + -. What is A in terms of x ? EXAMPLBthensz+4isdividedbyx‐1,theresultisA+;‐_7:‐1.WhatisAintermsofx?

x- 1

A)3x A ) 3 x-‐ 4

B)3x-3 B)3x‐3

C)3x C ) 3 x++ 3

D)3x+4 D)3x+4

Using synthetic synthetic division, division, Using

3x + 3 x X

3x22 + 4 -‐ 1 I3x 3x22 -‐ 3x 3x

3x + 4 3x ‐ 3 7 If you followed along, clunky when when we subtracted subtracted the the -‐ 33xx and and brought brought you followed along, you you should've should’ve noticed noticed it got got aa little little clunky the 4 down. That’s because because the the dividend, dividend, 3x 3x22 + + 4, 4, has has no no x term. process is the the down. That's term. Still, the the process the same: same: subtract subtract and and bring the remaining terms terms down. down. bring the remaining

164

THE COLLEGE PANDA PANDA THE COLLEGE

L1.

2 4 : 3x + + 7 The quotient quotient is 3x 3x + + 3 and the remainder 7. The The result result can as 33x +14 Now x + The and the remainder is 7. can be be expressed expressed as = 3x + 3 + _ __ Now x ‐- 1 x -‑ 1 it’s easy easy to see see that that A = = 3x 3x + + 3, 3, answer answer ~-(C) . it's

Again, we we could've could’ve done done thi thiss question question by by making making up up numbers. lf x = 2, then Again, = 2, then numbers. If 2 3x22 + +44 _ 3(2) 3(2)2+4 3x +4 16 - xx---‐ 1 1- = - 2 2---‐ 1 1- = 216

If we we didn't didn’t know know the the answer answer was was (C), (C), we we would would test test each answer choice with xx = 22 until 16, but since If each answer choice with until we we got got 16, but since we do do know, know, we'll we’ll test (C) first for confirmation. confirmation. Letting Letting A = = 3x 3x + + 3, we test (C) 3,

7 7 3X+3+m ‐3(2)+3+m ~ ‐99++77=: 116 6 + 3 + -= 3(2) + 3 + -= x- 1 2- 1

3x

Answer Answer confirmed. confirmed. 2

4

1

EXAMPLE 3: If the expression Sx - ~ + is written in the form Sx + 6 + __!!_, where B is a constant, EXAMPLEB:Hfl\eexpressim-5’:2‐Jt_‐__4ii‐liswritteniritheform5x+6+‐B-,whereBisaconstant, xx-‐ 2 x- 2 what is the value off B ? whatisthe value o Based division. Based on on where where it is, is, B represents represents the the remainder remainder of the the division. 5x 5 x+ +

x -‐ X

6

2 I5x 5x2‐ 4x 4x 2

+1 + 1

2 5x2‐10x 5x lOx

6 6x x ++ 1

6x ‐

12

13

13 13 r:;---:;-i Wecan can write write the the result result of this this division division as as5x x _ 2,, from from which = L.llJ. -. We Sx + 6 + -which B = 2 xThis last last example example is perfect perfect for demonstrating demonstrating aa shortcut shortcut called remainder theorem, get This called the remainder theorem, which which allows allows us us to get the remainder without without going going through through synthetic synthetic division. division. the remainder

In Example Example 3, 3, we we divided divided Sx 5x22 -‐ 4x 4x + + 11by 2. Whenever Whenever a divided by by a which is is 1n by xx -‐ 2. a polynomial polynomial is divided a monomial, monomial, which just something something in the the form form of ax ax + b, b, the remainder can can be be found the va value the remainder found by by plugging plugging in to the the polynomial polynomial the lue just that makes makes the the monomial monomial equal equal to to 0. The The process process sounds m o r e complicated than it is, so let’s of xx that sounds more complicated than is, so let's show show how how it’s done done.. it's What makes 0? x = z 2. What makes x -‐ 2 equal equal to O?

Plug xx = 22 into the polynomial polynomial 5x 5x22 -‐ 4x 4x + 1. 1. Plug into the

5(2)22 -‐ 4(2) + +11 = 13 13 5(2) A n d that's that’s the the remainder remainder we we obtained obtained in Example Example 3. And

165

DIVISION SYNTHETIC DIVISION 18 SYNTHETIC CHAPTER 18 CHAPTER

l? + 1? divided by x + - 2x 2 + when ‐23(2 remainder when the remainder What is the What + 5x is divided polynomial, the polynomial, into the Plugging that zero? x = -‐ 11.. Plugging to zero? equal to makes x + what makes Well, what + l1equal that into 7 ‐2(‐1)2+5(‐1) - 2(- 1) 2 + 5(- 1) = -‐7 the remainder - 7 is the Boom. ‐7 Boom. remainder.. l? divided by 2x 4x 4 + when 4x4 remainder when What + 3x 3x22 -‐ 4 is divided 2x -‐ 1? the remainder What is the 1 zero? x = equal to zero? makes 2x -‐ 1 equal What makes What z 2'.

2

polynomial, the polynomial, into the Plugging that into Plugging that

2 - 4 = 41+ 43- 4 = - 3 (21) + 3 (1) 4

4

2

remainder. the remainder. Boom . -‐33 is the Boom.

.· 2x2‐5x+1. .· m .· alent form 2x + 1 + -x-R R . ·. th e equiv 2x2x- _Sx + 1 IS· wntten uv"' 11.•PLE If th e expression __- , what is the EXAMPLEdzlftheexpreesmn‐‐;‐:§‐‐1swnttenmtheeqmvalentform2x+l+m,whatrsfire 3 3

J;,.IUUT-1,

A ,,.: .

value of R? valueofR?

plug in can plug we can theorem, we remainder theorem, the remainder Using the by x -‐ 3. Using R represents after dividing dividing 2x 2x22 -‐ 5x + 11by remainder after the remainder represents the 2 remainder. the get the remainder. x = 3 into + 1 to get 2x -‐ 5x + into 2x2

2(3)2‐5(3)+1=18‐15+1=. 2(3) 2 - 5(3) + 1 = 18 - 15 + 1 = [±J division . synthetic division. No need for synthetic No need Plugging in 2, 2. Plugging x 2 ‐- 3x + 2 by x ‐- 2. divide x2 we divide that we theorem . Let's remainder theorem. the remainder about the thing about One Let’s say say that 2, we we last thing One last see remainder is the remainder that the see that (2)22 -‐ 33(2) +2 =0 2z (2) + (2) factor we factor indeed, if we factor of 18. And a factor like 3 is a x 2 -‐ 3x + 2, just remainder is 0, x ‐- 2 is a factor of x2 the remainder Since just like And indeed, Since the 2 ‐ 3x + 2, x2 2, + 3x x 1) 2)( x ‐- 1) (x -‐ 2)(X (x that x -‐ 2 is in fact a factor. we see see that connected? math is connected? everything in math how everything Don’t love how just love you just Don't you 1? x 3 + 1? factor of x3 1 a factor + 1a example, is x + easier . For example, much easier. become much Some o w become questions nnow Some questions 3+ factor of xx3 Therefore, x + 1 is a factor = 0. Therefore, - 1) 3 + 1 = remainder is ((‐1)3 that the remainder Well, plugging 1. plugging in -‐ 11,, we find that

were we were If we like x + 1. monomials like dividing by monomials works when only works theorem only remainder theorem the remainder that the note that Do note when we're we’re dividing 1. If division . synthetic division. use synthetic have to use would have we would x2 + 2, we dividing like x2 something like by something + 11by x3 + dividing x3

166

PANDA COLLEGE PANDA THE COLLEGE THE

EXAMPLES: EXAMPLE 5:

+5Sxx++ 2 kx2 + 3x 3 ‐- kx2 f(x) = 3x3 f(x) =

k?? what is the 2, what by x ‐- 2, divisible by constant. If ff(x) a constant. is a above, kit is defined above, polynomial f(x) the polynomialf In ·the (x) defined (x) is divisible the value value of k 12 A) 12

B)9 B) 9

C)66 C)

0)33 D)

a words, x ‐- 2 is a other words, 2. In other divided by x ‐- 2. when ff(x) O when remainder is 0 the remainder then the 2, then If f (x) is divisible (x) is divided divisible by x -‐ 2, If f(x) zero) equal to zero) makes x ‐- 2 equal that makes va lue that (the value plug 2 (the we plug that when remainder theorem factor r ) The theorem tells us us that when we The remainder factor of f ((x). get 0. into (x), should get we should ), we into fJ(x

f/ ((2) 2 ) = 00 3(2)33 3(2)

‐k(2)2+5(2) 0 + 2 := o k(2) 2 + 5(2) +2 2 4- ‐ 4 1 0++22=: 0 4kk++ 10 24

-

36 -‐ 4k = 36 =0 ‐- 44kk := ‐- 336 6 k= 9 Answer Answer ~(B) .

EXAMPLE EXAMPLE 6: X

p(x)

-3

1

- 1

0

0

5

2

-3

4

4

the following Which of the values of x. Which some values p(x) for some polynomial p(x) value of polynomial the value gives the above gives table above The table The following must must be be p(x)?? a factor of p(x)

A ) x ++ l1 A)x

B)x B ) x- ‐ 1

C)x C ) x-‐ 4

D)x D ) x ‐- ‐ 5

Answer p(x). Answer factor of p(x). a factor be a must be = 0, x ++ 1 must p((-‐ 1) make s this theorem makes remainder theorem The remainder The this question question easy. easy. Because Because p l) = is NOT answer is Be careful-the answer NOT x -‐‐ 1. 1. ~-(A) . Becareful‐the

167

CHAPTER CHAPTER 18 18 SYNTHETIC SYNTHETIC DNISION DIVISION

CHAPTER EXERCISE:Answers for this chapter start on page 317.

A calculator should be used used on the should NOT N O T be the following following questions. questions. X

The expression x4_x x 2 is equa The expression equall to which which of the the x- 2 following? following?

3

6

0

A) A ) -‐ 2

8 B) ---+ x- 2

1

2 3 -4 -3

-3 -2

4

4

g(x)

The function g is is defined defined by by a a polynomial. polynomial. The The The function g(x). table above above shows shows some va lu es of x and some values and g(x). table What is the the remainder remainder when when g(x) g(x) is divided divided by What by xX + 33??

8 4 ) x -‐ 2 + - 2x 2x D) 44 ‐‐

C) --+ C ‐‐

A)) -‐ 2 A

B) 1 C) 2 2

. . th . 6x + 5x + 2 . IIff the expression M x + 1s th e expression is wntten written m in thee 22x + 11

__

' '

D) 6

7

1 Q, what is Q in terms of x ? form _2 xl_+ 1 + form +Q,whatisQ1ntermsofx. 2x + 1 A) A ) 33x x -- 1

2z3 - kxz2 + 5xz + 2x - 2 223‐kxzz+5xz+2x‐2

B) 3x + 1 C) C) 6x2+3x+1 6x 2 + 3x + 1

In the the polynomial polynomial above, is aa constant. constan t. If 2 z -‐ 11 above, kk is is a a factor factor of the the polynomial polynomial above, what is the is above, what is the

D) D) 6x2+5x+1 6x 2 + 5x + 1

value value of kk??

The written as The expression expression 4x 4x22 + 55 can can be be written as A(2x 1) + R, where A is an expression A(2x ‐ R, where A anexpression in terms terms of x and lu e of R and R R is aa constant. constant. What What is the the va value R ??

What is the th e remainder remainder when x 2 + 2x 2x + 1 is What when x2 is divided divided by by xx ++44 ??

168

THE PAN DA THE COLLEGE COLLEGE PANDA

A calculato calculatorr is allowed allowed on the following following question s. questions.

3 + x2 If p(x) = xx3 divisible p(x) = x2 ‐- 5x + 3, 3, then then p(x) p(x) is is divisible by which which of the the following? following? by

I. xX -‐ 2 When Bx -‐ 4 is by 3x -‐ 2, When 3x 3x22 -‐ 8x is divided divided by 2, the the result can be be expressed A -‐ result can expressed as asA

II. x ‐- l1 II. IIII. I]. x + + 33

x ~ . What What is A £133 A 2 3

A) A ) xx-‐ 4

and II only A) I and 11only and 11] III only only B) I and C) [I II and and IIII only C) l l only

B)) xx-‐ 2 B

D) I, I, II, II, and and III D) III

in terms of x ? intermsofx?

C) C ) xx ++ 2 D) D ) xx ++ 4

the polynomial polynomia l p(x) p(x) is is divisible divisible by by x ‐- 2, If the 2, p(x)?? the following could be be p(x) which of the which following could The The expression expression 2x 2x22 -‐ 4x 4x -‐ 33 can can be be written written as as A (x + where Bis constant. What What is A A(x + l1)) + + B, where B is a constant. in terms terms of x ?

= - x 2 + 5x - 14 B) p(x) = x 2 - 6x - 2 B) p(x)=x2~6x‐2 C) p(x) = 2x 2 + x - 8 C) p(x)=zx2+x‐8 D) p(x)= 3x2 - 2x - 8 D) p(x)=3x2‐2x‐-8

A) p(x))

A) 2x A) 2x + +6

B) 2x 2x + 2 C) C ) 22xx -‐ 2 D) D ) 22xx -‐ 66

If x ‐- 1 and and x + are both both factors factors of the the If + 1 are 2 4 3 po lynomial ax bx -‐ 3x and a and and bare polynomial ax4 + bx3 3x2 + 5x 5x and b are constants, what what is the the value value of a ?7 constants, The can be The expression expression xx22 + 4x 4x -‐ 9 can be written written as as (ax+ b, and and care (ax + b)(x b)(x -‐ 2) + + c, where where a, a,b, c are constants. What constants. What is the the value value of a + + b+ + c?

A ) -‐ 3 A) B) 1 C) 3

A ) -‐ 2 A)

D D)) 5

B) 3 C) 7

D D)) 1100

For a For a polynomial polynomial p(x), p(x), p

(5) of the (D= 0.0. Which Which of the

following following must must be be a a factor factor of p(x) p(x)?? A) 3x ‐- 1 l

(2) = 0. For a a polynomial polynomial p(x) For p(x),, pp(2) 0. Which Which of of the the folJowing must be true following m u s t be true about about p(x) p(x) ??

B + 1l B)) 3 3xx +

A) 2x is a a factor factor of p(x). p(x).

C) x - 3

p(x). B) 2x -‐ 2 is aa factor factor of p(x).

D) x + 3

C) x -‐ 2 is a a factor factor of p(x). p(x).

+ 2 is aa factor factor of p(x). p(x). D) x +

169

19

Numbers ComplexNumbers Complex 2 = imaginary the imaginary invented the mathematicians invented until mathematicians What value of x satisfies : -v1? There were w e r e no no values values until 1? There satisfies xx2 What value can power other from there, and from They defined ;=1. They represents \/‐1. number defined i2 i2 to equal equal -‐ 11,, and there, any any other power of i can which represents number i, which be derived. be derived.

;2 1 “ =: -7 1]

;3 =4-‐1i 1K ;4 14= z 11 1'5 L

i

;6 1"= :- fl1

;7 {7 =z -‐1i ;B i” = z 11 powers of i. For higher powers simplify higher i4 = that i4 the fact that repeat in cycles The cycles of 4. 4. You can can use use the : 1l to simplify For results repeat The results example, example, 1-511: (ml: X 1'2 ; 1 x i 2 : _] subtract, add, subtract, We add, complex number. a complex called a like 3 + When used in an an expression expression like + 2i, the the expression expression is called number. We When i is used expressions. algebraic would we like much numbers complex divide complex numbers much like we would algebraic expressions. and divide multiply, multiply , and equivalent to (3 the following EXAMPLE 1: If i = J=I, \/ ‐ 1 , which which of the following is is equivalent (3 + + Si) ‐- (2 (2 -‐- 3i) 3i) ?? EXAMPLE

A)9i A)9i

B )11+ +2 2ii B)

C C)) 1 + 8Bii

D) D ) 55 ++ 8Bii

Just expand and combine terms . like terms. combine like expand and Just ( 3+ + Si) 5 1 )- ‐ (2 ( 2 -‐ 33ii ) = 2 .3 + + 5Sii -- 2 + 3 : 11 ++8Bii 3i i = (3 Answer Answer ~(C ) .

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THE COLLEGE PANDA PANDA THE COLLEGE

A,

EXAMPLE 2: Given Given thati that i = what is the product (4 + ii))(5 EXAMPLEZ: = \/‐1,whatisthe product (4+ ( 5- ‐ 2i) ? A) 18 - 3i A)18‐-3i

B) 22 - 3i B)22‐-3i

C) C ) 18 1 8+ + 33ii -.

D) D ) 22 2 2+ + 3i 3i

Expanding, Expanding, ((4+i)(5‐2i):20‐8i+5i‐2i2=20‐3i+2=22‐3i 4 + i) (5 - 2i) = 20 - Bi + 5i - 2i2 = 20 - 3i + 2 = 22 - 3i Answer ~( B ) . Answer

.

.

.

22 + 3 i 1 +i

EXAMPLE the following is equal to ~i ?? EXAMPLE 3: Which Winch of the followmgisequal to 1+ +1.

1 1. A)§‐§l

1.. 1 1 B) B)§+'il -+- 1 2

2

5 1. (Di‐El

55 1. 1. D) D)§+El -+ -t 2 2

When you're with aa fraction containing i in the denominator, multiply multiply both and the the bottom When you’re faced with fraction containing the denominator, both the top top and bottom of the conjugate, conjugate, you conjugate of 1 + i the the fraction fraction by the the conjugate conjugate of the the denominator denominator.. What What is the you ask? Well, the the conjugate simply reverse reverse the sign between. is 1 -‐ i. The conjugate conjugate of 5 -‐ 4i is 5 + 4i. 41". To get get the the conjugate, conjugate, simply sign in between. In this top and this example, example, we multiply multiply the top and the the bottom bottom by the the conjugate conjugate 1 -~ i.

2 + i - 3i2 5 +i 5 +1 i. (2+3i).(1‐i)_2‐2i+3i‐3i2_2+i‐3i2_5+i 1

(2 + 3i) (1 - i) _ 2 - 2i + 3i - 3i2 (1 11-‐ i + i ‐- i ;2 ( 1+ + ii)) . (1 ( 1-‐ ii)) 2

_

11‐12 - i2

= - 22- = 2 + 2

The The whole whole point point of this process process is to eliminate eliminate i from the denominator denominator.. The absence absence of i in the the denominator denominator is aa good were done answer is ~(D) -. good indicator indicator that that things things were done correctly. The The answer

171

CHAPTER 19 CHAPTER 19 COMPLEX COMPLEX NUMBERS NUMBERS

CHAPTER EXERCISE: Answers for this chapter star t on page 319.

A calculator calculator should should NOT N O T be used on on the the be used following questions. following questions .

((66++2i)( fl fl 22+ + Si) 5) If + bi, bi, If the the expression expression above above is is equivalent equivalent to to aa + where a and b are constants, what is the value where and bare consta nt s, what is the va lue of aa??

For z \/‐_1, of the following is For ii = ;=T , which whic h of the following is equivalent Si)?? equivalent to (5 -‐ 3i) 3i) -‐ (( -‐ 22 + + Si) A)) 33 -‐ 8 Bii A

A) 2

B) 3 + + 2i

B) 12 12 B)

C) C) 7 -‐ Bi 81'

C)) 2222 C

D) + 2i 2i D) 77 +

D)) 3344 D

Which Which of the following is equall to the following is equa to

Given that that i = \/‐_1, Given of the the following following is is ;=T, which which of equal to ii(i ( i + 1) 1) ??

3(i0++2) 3 m-‐ 2(5 2 w- - 4i) 4 0?u(Note: w m ei =i =H¢ f fl)

A) i -‐ 2

A) 16 16 -‐ Si

B) i -‐ 1

B) -~44 + 7i

C) i + + 1l

C) -‐ 44 ++ 11i Hi

D) D) 0

D) 16 16 ++ 11i 11i

For i = = \/R‐ l ,, which For is which of the the following following is equivalent + 2) ‐ i ( i ‐ 1) ? equ iva lent to 3i(i 3i(i + 2) - i(i - 1) ?

fi+3fl+2 i4 + 3;2 + 2 Which of the the following Which following is equal equa l to the the expression expression above? (Note above? (Note:: i = H \/‐_1))

A A) ) ‐- 4 ++ 7ifi

A) i A)

B B) ) -‐ 22 ++ 7ifi

B B)) -‐ 1

C C)) -‐ 4 ++ 5ia D & D)) ‐- 2 ++ Si

C) ) 0 C

D) 1

For i = \/ R‐ 1 , which which of of the the following following is is equal equal to to ,-93 ? ? 193

22+m+43+5fi+6fl + 3i + 4i + 5{'- + 6i 2

4

A A)) -‐ 1

If the the expression expression above If above is equivalent to aa + + bi, bi, equiva lent to where a and where value of and b are are constants, constants, what what is is the the value of a+ +bb?? ((Note N o t ei i= = \/‐_1) H )

B B) ) 1

C)) -‐ ii C D D)) i

A)) 2 A

B) 6

C 0 C)) 110 D)) 112 D 2

172

THE COLLEG COLLEGE THE E PANDA PANDA

Which of the following complex numbers is is Which the following complex numbers equivalent (3 ‐ i )2? (Note: i = \/-‐‐1) equivalent to to (3 - i)2? (Note: i J=T) A) A)

Which of the following is equal to -1 -- 3i. ?? 3+ 1 (Note: i = (Note: = \/‐‐1) R )

88 -‐ 61" 6i

B) 8) 88 + 61' 6i

A A)) -‐ i ,-

C) 10 10 -‐ 6i C) 6i

B) 8) i

+

C)‐§i C) - ~i

+

D) 0)

10 10 + 61’ 6i

0) ‐‐§i ~ - ~i D) 4 4

(-i)2 - i)4 (‐i)2 -‐ ((‐i)4 number system, system, what what is is the In the the complex complex number the value (Note: = \/‐_1) value of the the expression expression above? above? (Note: ii = J=T)

Which the following Which of of the following complex complex numbers numbers is is 2 equivalent to 3‐1; equivalent : ‘/_1) - ~ ?? (Note: (Note: ii = J=T)

A)) -‐ 2 A

B) 8) 0

2+ 1

C) 1

3 4, - - l 2‐31“ 5 5 4, 8) 1 - - I 5 5 4. C) - - - l 3 3 4, 0) 1 - - / 3 A) A)

D) 2 0)

-

B)1‐§i

C)g‐gi

(5 (5 -‐ 2i)( 2i)(4 31') 4 -‐ 3i)

D)l‐§i

Which of the the following following is is equal equal to to the the expression Which expression above? above? (Note: (Note: ii = = \/‐_1) R )

_I__________ __,

A 4 -‐ 7 A)) 114 7ii

B) 14 -‐ 23i 8) 14 23i C) + 7i C) 26 26

4 + i + 2 ‐1 - i 1 ‐1 - i+ l + +1i

D) 26 0) 26 ‐- 231" 23i

Which to the Which of the the following following is is equal equal to the expression expression above? = \/‐‐1) above? (Note: (Note: ii = J=T) A) -‐22 -‐ i B) 2 + i 8)

1 11 1 Which of the following is equal to -:i‐2+i_4' +~ +~ ? I

(Note: i

I

I

C) 4 + i

= J=T)

D) 0) 4 ‐- i

A) - i 8)

C) 0 0)

1

173

20

Absolute Value Absolute The absolute absolute value value of x, denoted denoted by lxl, |x|, is the the distance distance xis x is from from 0. O. In other other words, words, absolute absolute value value makes makes The everything everything positive. positive. If it's it’s positive, positive, it stays stays positive. positive. If it's it’s negative, negative, it becomes becomes positive. positive.

EXAMPLE 1: How H o w many many integer integer values values of x satisfy satisfy lxl |x|

B

\

I

P

,,.;-.. ... ,.

D

I

Q C

lnflrefigumabave,fi,P§,and‐D'Careparakl. PointPliesonEandpointhiesonFC. lfB Q = 44,, In the figure above, AB, PQ, and DC are parallel. Point P lies on AD and point Q lies on BC. If BQ QC=2,andAD=7.5,whatisthelengthofAP? -QC = 2, and AD = 7.5, whatis the length of AP? Because E and W cut through through three three parallel parallel lines, lines, they they are are divided proportionally. Because AD and BC cut divided proportionally.

£_@-L2 PD_QC_2_

can see, since since BQ is twice twice QC, AP is twice twice PD (a ratio ratio of 2 to 1). A 2:1 ratio ratio means means that that AP is -2‐3‐1‐ : ~g - = As we can 2+1 3

2

the length length of AD AD.. AD AD is then then ~5 x 7.5 = : [fil. I. If you you prefer prefer to do length PD be x. the do things things algebraically, algebraically, then then let length PD be ‐

2

Then length length AP AP is 2x, and and since since AP AP and and PD PD sum sum to AD A D , 2x 2x + x == 7.5. This equation z 2.5 and we get Then equation gives gives x = and we

AP = : 2x = = 2(2.5) 2(2.5) = : @J I. . AP

196

THE COLLEGE PANDA THE COLLEGE PANDA

Radians radian is simply simply another another unit used to measure measure angles. have feet meters, pounds and A radian unit used angles. Just Just as as we we have feet and and meters, pounds and kilograms, we we have degrees and radians. kilograms, have degrees and radians. 7Tradians radians = 180° 180° Tr

If you've you’ve never never used used radians radians before, before, don don't be put put off it’s just just aa number. number. We We could’ve If ' t be off by by the the 7T. Tr. After After all, all, it's could've written written 3.14 radians radians :::::; z 180° 180C instead, but but everything typically expressed expressed in terms terms of 7r working with with radians radians.. Furthermore, Furthermore, Tr when when we're we're working instead, everything is typically is only only an anapproximation. given the conversion factor factor above, above, how radians? 3.14 is approximation. So, given the conversion how would would we we convert convert 45° to to radians?

7Tradians __ .:: 71 d" 450 45° xx Tr radians 180° 4 radians ra ians 1800 - Z Notice that that the degree units units (represented (represented by the the little little circles) cancel u t just just as they should should in any conversion Notice the degree cancel oout as they any conversion 3 problem.. Now N o w how how would would we convert convert 7” degrees? Flip Flip the the conversion ; to degrees? problem conversion factor.

3rr d.1ans x ---= -3‐H ra radians & 1800 : 2700 0° 2 Tr radians 7Tradians You might be wondering wondering why why we we even even need need radians radians.. Why n o t just just stick this another another difference might be not stick with with degrees? degrees? Is this difference between rest of the and meters? meters? Nope the chapter chapter on on between the the U.S. U S . and and the the rest the world, world, like like it is with with feet and Nope.. As we'll we’ll see see in the circles, some calculations are much easier are expressed expressed in radians circles, some calculations are much easier when when angles angles are radians..

EXAMPLE9: EXAMPLE 9: y

m

In the above, line origin and and has a a slope slope of y/j. the xy-plane xy-plane above, line m m passes passes through through the the origin Vi. IfIf point point A A lies lies on on line line m m and point point B lies lies on on the x-axis as as shown, what is the the measure, measure, in radians, angle AOB? AOB ? and the x-axis shown, what radians, of angle

A).:: A); 71:

6

B)?

n B) n: 5

C).:: D).:: C); D)? 4 3 7T

7T

197

CHAPTER 22 CHAPTER

TRLANGLES TRIANGLES

We can can draw from A to the We draw aa line line down down from the x-axis x-axis to to make make aa right the slope slope is v'3, V3, the ratio of the right triangle. triangle . Because Because the the ratio the height of this height base is always to 11 (rise (rise over over run). this triangle triangle to its base always \/5 v'3to run).

y m

This right right triangle triangle should should look triangle. Angle the fi, v'3,so This look familiar familiar to you. you. lt's It’s the the 30-60-90 30‐60‐90 triangle. Angle AOB A 0 3 is opposite opposite the so its its measure is measure is 60°. Converting Converting that that to radians, radians, 7T

60 x 180° 2 3

Answer Answer ~(D) .

198

THE COLLEGE COLLEGE PANDA THE PANDA

CHAPTER EXERCISE:Answers for this chapter start on page 324.

V.

calculator should should N NOT A calculator O T be on the be used used on following following questions. questions.

X

The lengths length s of the the sides sides of a The are x, a right right triangle triangle are 2, and and x + 5. 5. Which Which of the x -‐ 2, the following following equations could could be used to find x ?? equations be used A)

+X - 2 = X +5 + (x + 5)2 = (x - 2)2 + (x - 2)2 = (x + 5)2 (x - 2) 2 + (x + 5)2 = x2

X

square of side side length length 6 A square the figure 6 is shown shown in the figure above. What the value above. What is the value of x ??

B) x2 C) x2 D)

6

A) WE A) 3v'2 B) 6

C) WE C) 6v'2 D) 6v'3 6\/.’§ B

B

I C A D Note: Note: Figure n o t drawn drawn to scale. Figure not ln BDC above, the length In .6 ABDC above, what what is the length of W DC?? C

A) 3

the figure figure above, above, E In the || CB. What is the AB II CD. What the length length of AB AB??

B) 5

C) 5\/§ C) 5v'3 D) 8

199

CHAPTER 22 TRIANGLES CHAPTER TRIANGLES

Two angles same measure. measure. angles of a a triangle triangle have have the same IfIf two 15and the two sides sides have have lengths lengths 15 and 20, what what is is the greatest possible greatest value of the of the possible value the perimeter perimeter of the triangle? triangle?

Note: drawn to scale.. Note: Figure Figure not not drawn to scale

In the the figure figure above, above, the the base base of of a a cone cone has has a a radius sliced horizontally so radius of 6. The The cone cone is is sliced horizontally so that the top that piece is a smaller cone with a height top piece smaller cone with height of 1 and and a base base radius radius of 2. What What is the the height height of of . the bottom the bottom piece? piece?

N N

~ 5

M

5

8

A) 1

B) 2 C) 3

O 0

D) 4

What area of isosceles What is the area M NO isosceles triangle triangle MNO above? above?

A

B

H

10

C

Note : Figure not drawn to scale . the figure In the figure above, above, E AB is parallel parallel to to flGH and and 57 OF is parallel ‐C". If = 1, 1,EH = 3,EC 2 2, 2, and parallel to 'BBC. If DE DE = EH = 3, EG = and HC = 10, what what is is the length OfAD of AD?? HC 2 10’ the length

the figure figure above, In the anequilateral on above , an equilateral triangle triangle sits on top of a a square. top If the an area of 4, square. If the square square has has an area what is the what the area area of the the equilateral equilateral triangle? triangle? A)) \ v'3 A /§

\/§ B) )v'3 B T 2 3 OZ C) 4 D D) ) 1

200

THE COLLEGE COLLEGE PANDA PANDA THE

A calculator is allowed allowed on the following following questions questions..

cC

DE

How radians are H o w many many radians are in 225° ??

A)

37T 37r -

D...,______

T4

7rr 77r B) -

?6

5rr Sn C) -

A

74

0)

E

B

ABC above, L CDE = and AA L A = 90° In 6 AA B C above, ACDE = 90° and 90°.. AB : 9and DE = 6, 6,whatis AB = 9 and AC AC = 12. If DE what is the the length of CE length CE??

337r rr -

72

A)) 6 A

B) 8 C) 9 0)) 110 D 0

A

~

B

9

C

E

F

w W

Triangle ABC ABC above similar to to triangle triangle DEF. DEF. Triangle above is similar What is the perimeter of triangle DEF ? What the perimeter triang le DEF? A ) 220 0 A)

zZ 20 20

B) 26.8

C C)) 3300

12 12

D) 36.2 0) ________________

X

J

15 15

yY

Two above are are Two poles poles represented represented by by W XW and and W YZ above 15 feet 20 feet 15 feet apart. apart. One One is 20 feet tall tall and and the the other other is 12 feet rope joins joins the 12 feet tall. tall. A rope the top top of one one pole pole to the the top of the other. What is the length of the rope? top the other . What the leng th the rope?

ABC,, E shortest side side.. In isosceles isosceles triangle triangle ABC BC is the the shortest If the degree measure of LA is a multiple of 10, the degree measure L A is a multiple 10, what is the the sma smallest possible measure measure of L [BB ?? what llest possible

A) 75°

A 2 A)) 112

B) B) 70°

B) B) 17 17

C) 65°

C) C) 18 18

D) 60° 0)

D) 0) 19 19

201 201

CHAPTER 22 TRIANGLES CHAPTER 22 TRIANGLES

y

A( - 2,4) 24

35 What is the What the perimeter perimeter of the the trapezoid trapezoid above? above? A) 100 A)

B) 108 C) 112 D) 116

B(- 2, - 3)

C(5, - 3)

Points A, B, and C form a triangle in the xy-plane shown above. What is the measure, in radians, of angle BAC?

A);6 7[

A)

B)? B) 4 C)? C) 3

8

7[

7[

X

D); D) 2 7[

What is the the value triangle above? What value of x in the the triangle above?

y

A

z

D

parallel lines lines are Two parallel are shown shown in the xy-plane the xy-plane above. If If AB AB := 15 and point point B has has coordinates coordinates above. 15 and n ), what what is the the value (m, n), value of n? n?

ln the the figure figure above, above, ABC ABCDD is aa square In square of side side length 3. AW = length 3. If AW : AZ AZ = CX = CY : 1, 1 , what w h a t is is CY = the perimeter rectangle WXYZ the perimeter of rectangle WXYZ? ?

A A)) -‐ 66

A s fi A) ) 3v2 B) B) 4\/§ 4v2 C e fi C) ) 6v2

B B)) -‐ 8 C)) -‐ 9 C

D 2 D)) -‐ 112

D) 8

202

THE COLLEGE PANDA THE COLLEGE PANDA

B

A

B

D

C

In the figure above, circle O is inscribed in the square ABCD. If BO = 2, what is the area of the circle? In the figure above, equilateral triangle A BC is inscribed in circle D. What is the measure, in radians , of angle ADB?

A) A)

7T

A)

4 7T

27T 27r

?3

37T 37r 8) B) -

8)

2

C)

7T

D)

T4

37T 2

47T 471 C) C) “5‑ 5 57T Sn D) D)

76

y

D

A

X 2

In the the figure figure above, [n above, the the value value of % where ~~ is k, where

. C

constant. Which Which of the the following following ratios k is a constant. ratios has has 1 a a value value of ik??

45° B

What the length length of DB the figure figure above? What is the DB in the above? A)

2v'3 M5 3 3

8) B)

2./6 216

C) C)

4./6 sfi 3 3

z Z

3

YZ

A) A)

flxz

8) B)

W xw

C) C)

W XY

xy XY yz YZ

YW YW

D) D) XW xw

\/5 D) v'3 203

CHAPTER 22 TRIANGLES CHAPTER TRIANGLES

y

B

B(2,8 ) E

F

+---+--------'~--x 0

A

D

C

0 (8,0 )

In the xy-plane above, points A and C lie on OB and BO, respectively . If AC is parallel to the x-axis and has a length of 3, what is the length of

Equilateral triangle triangle DEF DEF is inscribed inscribed in __ Equilateral equilateral triangle triangle ABC ABC such that E equilateral such that 1 AC. ED 1AC. What is the the ratio ratio of the the area area of ADEF 6. 0EF to the the area area What ABC?? of ABC

BC?

A)) 11 ::4 A 4

B)) 1 ::33 B 1:2 C) 1 :2

0)) 5 ::88 D

v

B

C

A

D

A

A‘ B

the figure figure above, above, equilateral equilateral triangle triangle AED In the AED is

the figure figure above, above, a a semicircle semicircle sits sits on on top top of a In the a square of side side 6. Point Point A is at the the3p top of the the square semicircle . What What is the the length length of AB? semicircle. AB ?

contained within within square square ABCD the contained ABC D.. What What is the degree measure measure of [LBBEC? degree BC?

A) B) C) D)

60° 100° 100° 120° 120° 150°

A) A) 3\/§ 3 v15 B) 7 C) 9

D 3m D)) 3v'IO

204

THE COLLEGE COLLEGE PANDA THE PANDA

In AABC, ZABC = 120° 120°.. 6 ABC, AB AB = BC BC := 6and 6 and L ABC = What is the the area area of AABC 6 ABC?? What

A

A N5 A)) 2v'3 B) B) 4\/§ 4v'3 C) 6v'3 C ) s fi D) W3 D) 9v'3

3

0 O

4

E

C

In the square DBCE DBCE has has aa side side the figure figure above, above, square length 4, what is the length AD ? length of 3. If OE OE = 4, what the length of of AD?

y 3!

e

x ( v'3, -‐ 11)) N3,

A

In the xy-plane above, above, angle formed by the xy-plane angle 9 0 is is formed by the the x-axis and and the the line shown.. What line segment seg ment shown What is the measure, in radians, measure, radians, of angle angle 0 ? ?

?3

B) B)

T4

-‘ 4 4

F

571 57T

A) A)

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205

CHAPTER TRIANGLES CHAPTER22 22 TRIANGLES

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206

23 Circles Circles

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Using these these values, values, we can Using can n o w construct boxplot: now construct the the boxplot:

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248

THE COLLEGE PANDA THE COLLEGE PANDA

CHAPTER EXERCISE:Answers for this chapter start on page 338.

A calculator is allowed allowed on the following following questions. questions.

The average height students in one The average height of 14 14students one class class is is 63 63 inches inches.. The The average average height height of 21 21 students students in another classes are another clas classs is 68. If If the the two t w o classes are combined , what the average inches, combined, what is is the average height, height, in inches, of the the students students in the the combined combined class? class?

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Kristie taken five tests Kristie has has taken tests in science science class class.. The The average of all test scores average all five of Kristie's Kristie’s test scores is 94. The The average average of her her last last three three test test scores scores is 92. What s? What is the the average average of her her first first ttwo w o test test score scores?

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A food food company company hires hires an an independent independent research research agency agency to determine determine its its product's product’s shelf shelf life, life, the the length length of time time it may may be be stored stored before before it expires. expires. Using random sample units of the Using aa random sample of 40 40 units the product, the the research product, research agency agency finds finds that that the the product's s. Which product’s shelf shelf life life has has aa range range of 33 day days. Which of the st be the following following mu must be true true about about the the units units in the sample? sample? the

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A) All All the the units units expired expired within within 33 days days.. B) The unit unit with with the the longest longest shelf shelf life life took took 33 B) The days unit with the days longer longer to expire expire than than the the unit with the shortest shelf shelf life life.. shortest

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249

CHAPTER STATISTICS I CHAPTER 27 STATISTICS

Locks are are sections sections of canals canals in which which the the water water level to raise level can can be be mechanically mechanically changed changed to raise and and lower shows the number lower boats. boats. The The table table below below shows the number of locks locks for 10 canals canals in France. France .

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shoe store store surveyed surveyed a a random random sample A shoe sample of 50 50 customers to better better estimate estimate which which shoe shoe sizes customers sizes should be be kept kept in stock. stock. The The store store found found that that the the should median shoe shoe size size of the the customers customers in the the sample sample median inches. Which Which of the the following following statements statements is 10 inches. must be must true? true?

The above gives the distribution The table table above gives the distribution of low low temperatures for a a city over over 28 28 days. days . What the temperatures What is the median low temperature, temperature, in degrees degrees Fahrenheit Fahrenheit median low (°F), of the the city for these these 28 28 days? days? (°F),

The sum sum of all the the shoe shoe sizes the sample sample A) The sizes in the inches . is 500 inches. average of the smallest shoe size and and B) The average the smallest shoe size the largest largest shoe the sample sample is 10 the shoe size size in the 10 inches. inches. C) The The difference difference between between the the smallest smallest shoe shoe C size and and the the largest largest shoe shoe size size in the the sample size sample inches. is 10 10inches. D) At At least least half half of the the customers customers in the the sample sample D have shoe sizes greater greater than than or equal equal to 10 have shoe sizes 10 inches. inches.

250

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The The tables tables below below give give the the distribution distribution of travel travel times times between between ttwo w o towns towns for Bus Bus A and and Bus Bus B 40 days . over over the the same same 40days. Bus Bus A Travel Travel time time (minutes) (minutes)

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The bar bar chart above shows the distribution The chart above shows the distribution of weights (to (to the the nearest pound) for for 19 kayaks weights nearest pound) kayaks made and 19 kayaks kayaks made made by made by by Company Company A and by Company B. B. Which of the the following Company Which of following correctly correctly compares the median median weight the kayaks compares the weight of the kayaks made made by each each company? company?

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Which Which of the the following following statements statements is true true about about the the data data shown shown for these these 40 days? days?

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A) The The standard standard deviation deviation of travel travel times times for for Bus Bus A is smaller. smaller.

B) The The median median weight weight of the the kayaks kayaks made made by by Company B is smaller. smaller. Company

B) The standard standard deviation deviation of travel travel times times for for B) The Bus Bis B is smaller smaller..

median weight kayaks is C) The The median weight of the the kayaks is the the same companies . same for both both companies.

C) C) The The standard standard deviation deviation of travel travel times times is the the same same for Bus Bus A A and and Bus Bus B.

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the provided . the data data provided.

Quiz Score

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6

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The above shows shows the the scores scores for Jay's first The table table above for Jay’s first seven quizzes . Which the following following are are seven math math quizzes. Which of of the true about his his scores? true about scores?

1, The The mode mode is greater greater than than the the median. I. median . n. The greater than than the the mean. mean . II. The median median is greater III. The range is greater than 20. The range greater than

A) B) C) D)

251

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2 1 222 2 23 23 2 2 5 26 26 2 2 8 29 2 9 30 30 21 244 25 277 28 Gas mileage mileage (miles (miles per per gallon) gallon) Gas The dotplot dotplot above above gives gives the the gas gas mileage (in The mileage (in miles per per gallon) gallon) of 15 15different cars. If the the dot dot miles different cars. representing the the car car with with the greatest gas gas representing the greatest mileage is removed from the dotplot, dot-plot, what what will will mileage removed from happen to the the mean, mean, median, median, and and standard standard happen deviation deviation of the the new n e w data data set?

Number of lectures lectures

Number of professors professors

12 15 21 25 28 32 40

15 12 6 20 17 15 5

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The summarizess the distribution of The table table above above summarize the distribution the number of lectures the number lectures each each of the the 90 90 professors professors at last year. year. Which Which of the at acollege a college gave gave last the following box plots data following box plots correctly correctly represents represents the the data shown in the the table? shown table?

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• Statisti Statistics II cs II The goal goal of statistics statistics is to be be able able to make make predictions predictions and information.. and estimations estimations based based on on limited limited time time and and information might want the mean mean weight of all all female female raccoons raccoons in in the United For example, example , aa statistician statistician might want to estimate estimate the weight of the United States. to survey survey the the entire raccoon population population.. In In fact, fact, by by the States . The problem problem is that that it’s it's impossible impossible to entire female raccoon the time be accomplished, accomplished, not n o t only only would would the the data data be u t of would be e w females in time that that could could be be oout of date date but but there there would be nnew female s in the population. population. Instead, Instead, aa statistician statistician takes random sample sample of female make an an estimation of the takes aa random female raccoons raccoons to to make estimation of what the actual actual mean mean might might be be.. In other words, the the sample sample mean the population population mean mean.. what the other words, mean is is used used to to estimate estimate the Using a sample to predict predict something o m m o n theme theme in in stati statistics and in in SAT SAT Using a sample something about about the entire entire population population is is aa ccommon stics and question s. questions.

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liters liters per per minute minute..

Using a prediction dangerous, especially especially when Using the the line line of best best fit to to make make a prediction can can be be dangerous, when

we are are making making aa prediction prediction outside outside the the scope scope of oour u r data oxygen uptake uptake at heart rate rate •0 we data set (predicting (predicting the the oxygen at aa heart le-you'd probably dead). of 250 beats beats per per minute, minute, for for examp example‐you’d probably be be dead). there are are outliers outliers that may hea heavily influence the the line Part 2). •0 there that may vily influence line of best best fit (see (see Part

•o the curve rather than aa linear one . In this this case, case, a a the data data is is better better modeled modeled by by aa quadratic quadratic or or exponential exponential curve rather than linear one. linear like compound compound interest linear at linear model model looks looks to be be the the right right one, one, but but something something like interest may may look look linear at first even even though it's it’s exponential exponential growth growth.. though Part 22 Solution: data point point farthest farthest away from the best fit is at at Solution: From From the the scatterplot, scatterplot, we we can can see see that that the the data away from the line line of best

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liters per minute . 118 along uptake of 2.s liters along the the x-axis. x-axis. The The point point represents represents an an oxygen oxygen uptake per minute. Note point is likely an hea vily influence the line our Note that that this this data data point an outlier, outlier, which which can can heavily influence the line of best best fit and and throw throw off o ur predictions . Outliers if they they represent represent specia cases or exceptions. exceptions. predictions. Outliers should should be be removed removed from from the the data data if speciall cases Not the line line of best best fit, but but you ' ll also also be be asked interpret N o t only only will will you you be be asked asked to make make predictions predictions using using the you’ll asked to to interpret its We'll use example in the the next next one one to show these concepts its slope slope and and y-intercept. y-intercept. We’ll use the data data from from this this example show you you how how these concepts are tested. are tested .

255

CHAPTER CHAPTER 28 STATISTICS STATISTICS II II

EXAMPLE 3: Exam 3: Oxygen Uptake versus Heart Rate

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dllring Kyle's exerciseroutine. The vdminglfiyle‘sexemisemtine. Thelineothestfitisalsoshown. line ofhest fit is al.so shown.

PART'l:Which, of th following is the best-interpretation of the slope .of the line of best fit in the context “M1:Winchofthefollowmgisthebestintapretahonotflieslopeoffllehneofbestfitinthecontext ofthispmblem? of this problell\'? A) ‘I'hepredrctedmcreasemele’soxygenuptake inlitersperminute, foreveryonebeatperminute The predicted increase in Kyle's oxygen uptake, in liters per minute, for every one beat per min~te in his heart,:rate ' increase incrmeinhishemme

B) The predicted B) The predicted increase increase'in in Kyle's Kyle's heart heart rate, rate, in beats beats per per minute, for every liter per per minute minute minute, for every one one liter

jnctease oxygen uptake increasem in his his oxygen uptake ‘ C) Kyle's Kyle'spredicted oxygen uptake uptake in per minute minute at heart rate rate of beats per minute predicted oxygen in liters lifers per at a a heart of 0 Obeats per minute predicted heart heart rate rate in in beats beats per per minute minute at at an an oxygen uptake of liters per D) Kyle’s Kyle's predicted oxygen uptake of 0 Oliters per minute minute

PART Which of the following is the the best best interpretation interpretation of y-intercept of of the the line line of fit in in the the PART 2: Which the following of the they-intercept of best best fit commit of this problem? context-of this problem?

A) The predicted increase in Kyle’s oxygen uptake, uptake, in in liters liters per minute, for every one beat beat per minute minute A) The predicted increase in Kyle's oxygen minute, for every one increase his heart increase in his rate heart rate B) the 13) The _pre,djcted.increase predictedi n c r e a s e in in Kyle's Kyle’s heart heart rate, rate, in in beats beats per per minute, liter per per minute minute minute, for every every one pne liter increase :m,his increasein his oxygen oxygen uptake ptake C) Kyle's predicted predicted oxygen uptake in in liters liters per per minute minute at heart rate beats per minute C) Kyle's oxygen uptake at a a heart rate of 0 0 beats per minute Kyle’s predicted predicted heart rate in in beats beats per per minute minute at at an an oxygen minute D) Kyle's heart rate oxygen uptake uptake of of 0 0 liters liters per per minute Part 1 Solution: Solution: As we we learned model questions the interpretation chapter, the slope is is the the learned in the the linear linear model questions in the interpretation chapter, the slope increase uptake) for each increase in x (heart rate). The only difference n o w is that it’s a predicted increase in y31(oxygen (oxygen uptake) each increase (heart rate) . only difference now that it's predicted

[0]

increase. increase . The The answer answer is (A) .

Part 2 Solution: Solution: They The y‐intercept the value value of y (oxygen (oxygen uptake) uptake) when answer is -intercept is the when xx (the (the heart heart beat) beat) is is 0. 0. The The answer is

-. Note that value would would have have no significance in in real real life since heart rate rate of of 0. [@. Note that this value no significance since you you would would be be dead dead at at a a heart 0. This again again illustrates illustrates the danger danger of predicting values outside outside the scope sample data. This predicting values scope of of the the sample data .

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EXAMPl.E4: M~d~~ a town; th~stateoi.M~chuse~ . A'rea;t\i stateagefifjandbnµy apartm~ts for .s~e m¥a:tden ~d foung that the a~f~ gepri ~eof ~j cl! ~~~ t .wa_s f 150A~2_2_15=0

V + 44 =z 99 Jxx + xX + 44 ==881 1 xX ==777 7

A2‐2A‐15=0 A2 - 2A - 15 = 0 ( A -‐ 55)(A ) ( A++ 3) 3 )== 0 (A 0 A= = 5, -‐33 xX xX .. .. So = -‐ 33.. Solvmg Solvmg these So 8 := 5 or E = these equations equations

14. (E] 14.‑

6

2 20‐fi=§fi+10 20 - ..rx = Fx+10 3

18. [zJThere w o ways There are are ttwo ways to approach approach this this problem. The problem. The faster faster way way is to factor factor the the

5 102§fi 10= 3../x 3

numerator numerator first. first.

e fi 6 =zv'x 3 = Xx 366=

2 xx2_ - 4x+3z4 4x + 3 = 4 x -‐ 1l

(x‐3)jJ//1‘)':4 3 )__(,x.----lj =4

Square both both sides sides of the 15. 11] Square the equation. equation.

(X -

M .x--T

2

x -‐ 3 = : 4

+ y)2 = (Jx2 + (x +y)2 = («33 +y2 y2 +16) + 16)2

x= =7

x2+2xy+y2 x2 + 2xy + y2=x2+y2+16 = x 2 + y2+ 16 2xy 2xy = 16 xy = = 88 xy 16.

6

gives x = 30 gives 8 . Since the 30 and and x = -‐ 118. the question question specifies that specifies that x x > 0, the answer answer is 30.

The second second way The r i d of the the fraction way is to get get rid fraction by multiplying multiplying both first by both sides 1 and and sides by x x -‐ 1 then then factor factor later. later.

2 - 4x + 3 _ xx2‐4x+3 =4 x -‐ 1 4 x2‐4x+3:4(x‐1) x2 - 4x + 3 = 4 (x - 1)

[g Cross both sides. sides. Cross multiply multiply and and expand expand both 4(2x‐ 1 )z ) ( x-‐ 22)) 4(2x - 1) = ((xx ++ 22)(x

8 : xx22 -‐ 44 Bxx- ‐ 44= 0=x2‐8x 0 = x 2 - Bx 0 0= = xx((xx -‐ 88))

x2‐4x+3=4x‐4 x 2 - 4x + 3 = 4x - 4 x2-8x+7=0 x 2 - Bx + 7 = 0 (( Xx -‐ 77))((Xx- ‐ 1 =0 1)) =

xX ==00,8 ,8

xX ==77,1 ,1

Neither are false solutions, Neither are set solutions, so so the the solution solution set is is {0,8}.

Now, Now, x = solution since since it causes = 1 is a false solution causes division by 0. Therefore, division = 7. Therefore, the the solution solution is x =

290

THE COLLEGE PANDA COLLEGE PANDA

calculator question, no calculator a no this is a Because this 19. [}] Because question, can strategy. You can valid strategy. a valid check is a and check guess and guess following: the following: do the also do also

23.

II] m2g ‘- µm1g = "128 a= l e g m1+ "12 + m2 m1

x20:4 9) = 8x 8x44 x2 (x4 ‐- 9)

a(m1+ ymlg m2g ‐- µm1g = ng a(m1 + m2) = = ‐- yµm1g m2g = a(m1 + m2) ‐- "128 a("11+ m1g

x - 9x - 8x = x6‐9x2‐8x4 :00 2 - 8x -‐ 9) = x 2 (x 4 ‐8x2 x2(x4 z 00 6

2

4

ng a(m1 + m2) m2g ‐- a(m1+ m1g m lg

x2(x2‐9)(x2+1)=0 x2 (x 2 - 9)(x 2 + 1) = 0 x2(x‐+‐3)(x 0 1) = 0 3)(x 2 + l) x 2 (x + 3)(x ‐- 3)(x2

24.

the out, the cancel out, 2's cancel the 2’s Because the [JJBecause same. the same. stays the acceleration stays acceleration

equation the equation be 3 for the must be 0, x must Because Because x > 0, true. be true. above to be above

_ 2 n g ‐ y(2m1)g = 207123 ‐ umig) 2m1+2m2 20711 + 7712)

y values. x and J/ given 20. II] 20‘ C Plug Plug iinn tthe h e 81 ‘venxand values.

anew ‑

:=” oaold ld

kx 2 + 5 y + 2kx = kx2

23+ = k(3) k(3)22 + 5 2(k) (3) = 23 + 2(k)(3) 2 + 6k 6 k== 99kk+ 5 233+

25.

‐- 33kk = ‐- 118 8 k =: 6 21.

E_x+12 X + 12 X

26.

42

Then side. Then hand side. left hand the left expand the First, expand First, [}QJ

combine factor . and factor. terms and like terms combine like

4 2 x=: 66xx ++7722 42x 3 6 x = 7722 36x x= : 2

x6

[]J 3(x 3 ( x-‐ 22y) y ) -‐ 33z2 = : 00 3 y --3 2 3z == 0 3xx -‐ 66y 3x 6y+32 + 3z 3x =: 6y xX = = 2y 2y+z +z

multiply . Cross multiply. II] Cross 6

= fl

((xx+ + ll ))((xx-‐ 22)) = 77xx -‐ 118 8 2 - x - 2 = 7x - 18 xxz‐x‐2=7x‐18

6

x2‐8x+16:0 x 2 - Bx+ 16 = 0

2

(x‐4)2=0 (x - 4)2 = 0

N o w , =‐ ‐ ‐ == 3. 3 Now,

22.‑ 22. II]

Therefore, and Therefore, x = 4 and 7x ‐ 18= 7(4) ‐ 18= 18 = 10. 7x - 18 =

c+1

d = a( c~ l) d=“< 24) a C+ l) =a(c~ i§=“(z_4>

1

cC + 11

5 2 : 247 1 2=: cC + + 11 12 l11l = z cC

291 291

CHAPTER CHAPTER 30 ANSWERS ANSWERS TO THE EXERCISES EXERCISES

IRJ

27. El We can can either answer choices either plug plug in the the answer choices or solve solve algebraically. algebraically. Plugging Plugging in the the answer answer choices choices is more more efficient efficient here, here, but but since since that’s that's self‐explanatory, solve algebraically. algebraically. self-explanatory, let’s let's solve First, both sides. First, square square both sides.

is false. A.. For choice B, B, false. We can can eliminate eliminate A For choice 2,/2 = Js, so the value 2\/2 = \/§, so ifif that's that's the value of Jx \/x -‐ 10, then have to equal we plug then x would would have equal 18. If we plug 18 into into the the equation, equation, the the left hand hand side x = 18 side is \/18 10= the right right hand side is is J 18 -‐ 10 = 2\/2 2 ,/2 and and the hand side m‐ /18 - fz=3\f2‐ ,/2 = 3 ,/2 - « ,/25=: 2NE. ,/2 . Sinceboth Since both sides be the the answer. sides match, match, choice choice B must must be answer .

(2/x)2 = (x - 3)2 (2\/§)2:(x‐3)2

4x=x2‐6x+9 4x = x 2 - 6x + 9

30.

IT]

O=x2‐10x+9 0 = x 2 - l0 x + 9 = ((xx ‐- 1l ))(x( x -‐ 99)) 0=

xy2+x‐y2‐1=0 xy 2 + X - y2 - 1 = 0 (y2+ 1) - (y2 + 1) = 0 X(y2+1)‐(y2+1)=0 (y2+ 1) (X - 1) = 0 (y2+1)(x‐1):0

X

=

x = 11,9 ,9 X Since 1 is a a false solution, the the only Since false solution, only value value of x that satisfies satisfies the the equation that equation is 9.

Since y2 Since + 11is always positive, equal y2+ is always positive , x must must equal 1.

28. 28.[Il

31. 4 4

_ = xx 22- ‐ 6 6xx++99 _

44

(x - 3) 2 (Y‐3)2

9

[I]

Divide both sides Divide both sides by P and and take take the the tth tth root of both root both sides. sides. V = P((11- ‐ r) : P r)’1

= =99

~p = (1 -‐r)1 r) 1

g=(x‐3)2 ; = (x - 3) 2

fig: ± '9= / (x-

~=

1- r

(x‐3)2 3) 2

,IV

r = l - Vp

i§=x‐3 3 2

±- = x - 3

32. 29.

I]] We can can use use the the answer answer choices choices to

[I]

From the the previous From question, we we know know previous question,

_

tthat h a rt r=‐ 11‐-

backsolve or we can backsolve can solve solve the the equation equation algebraically. We'll first solve solve it algebraically algebraically. algebraically by squaring squaring both both sides. sides.

,IV

i/V zP =‐_12.. $.BecauseVishalfP, V p· Because Vis- half P, p 2

Thus,r:1‐{‘/gz0.l3 = 1 - if"{:=:::: 0.13

Thus , r

(✓x - 10 ) 2 = (vx - 12)2 (x/x‐10>2=(f‐\/§)2

(Jx)2 /x)(n /h5 ) + ( /2 fi f -‐ 2( aw x / )2 if x‐ 10= :x -x‐zx/fln x - 10 2../h + 2

xx -‐ 10 1 0= :

-‐12=‐2\/fl 12 = - 2./ix 6 = ...fix, 6=\/2_x

3 6=: 22xx 36 1 8== Xx 18 Let's we wanted wanted to test Let’s say we answer test the the answer choices instead . H How choices instead. o w would that? For would we do that? For choice ../6is the the value choice A, if if \/5 \/xx -‐ 10, 10, then then x value of J would have have to equal equal 16. If we we plug z 16 16 would plug x = into the equation, equation, we we get into the get 44 = = 4 -‐ \/2, ,/2, which which

292 292

V

1

THE THE COLLEGE COLLEGE PANDA PANDA

Chapter 9: More Equation Strategies Equation Solving Solving Strategies CHAPTER EXERCISE: EXERCISE:

1. I55 I30 30 ((x3 + 1. x3 +

+ ~§x> ~éxz x2 + x)

haven’t assumed that haven't assumed that a and and b are are positive positive here. here . But of the the answer answer choices, choices, only only 6 is aa possible = 3 and and bb = = 22 or possible value value of ab ab (when (when a = when z -‐33 and = -‐ 22). ). when a = and b =

= 30x 30x33 + + 5x 5x22 + + 20x. = 20x.

Therefore,aa = 30, 30,bb = 5, and z 20. Therefore, and c =

+ bb ++cc = : 555. 5. a+ 2.. [Qj E] For For an to have an equation equation to have infinitely infinitely many many solutions, sides must must be solutions, both both sides be equivalent. equivalent.

5.

0

Expand both sides of the equation. 3 2 1 -- x - 4 = 4 -2

Comparing sides, ~a ga = ~g Comparing the the terms terms on on both both sides, and 3 = = 9b. Solving Solving these these equations, equations, we we get get and 1 a 4 a= = 4 and Therefore, b 5 = I= 12. and b = 5.. Therefore,

3

‐ Ex‐ 4‐ ‐ 2x

3

Let’s rearrange the right side that the the terms terms Let's rearrange the right side so so that line left side. side. line up with with the the left

3.. [Qj [E For For an no solutions, solutions, the an equation equation to have have no the coefficients coefficients of the u s t be be the same the x terms terms m must the same on either side but m u s t be be either side but the the constants constants must different. If we different. we expand expand the the right right side, side, we we get get

1 1 -‐ ‐xx -‐ 4 =-: ‐ ‐xx+4 4 2 2 +

Now it’s Now easy to see see that that the the it's easy the coefficients coefficients of the x terms terms are same but the constants constants are are are the the same but the different the equation different ( -‐ 4 vs. 4). Therefore, Therefore, the equation

a axx-‐ b = = 66xx + 3 Therefore, aa = = 6 and and b f= 742 ‐ 33.. Only Only answer answer Therefore, choice D satisfies satisfies these these conditions. conditions. Choice choice Choice C would result in infinitely many solutions solutions would result infinitely many since both both sides would be beequivalent. since sides would equivalent.

has solutions. has no no solutions.

6.

0

= (p (F7 + + q)(P Q)- We 4.. Remember,P2 Remember, p2 -‐ £72 q2 = q)(p -‐ q). We can take can apply apply this this factorization factorization here here once once we we take out aa 2: 18x22 -‐ 8 = 2(9x2 = 2(3x ++ 2)(3x 2)(3x -‐ 2), 2), 18x 2(9x 2 -‐ 4) = which equals 2(ax 2(ax + b)(ax b)(ax -‐ b). b). Comparing Comparing which equals the coefficients, coefficients, a = 3 and and b = : 2. Therefore, Therefore, the ab = : (3) (3)(2) : 6. Note that this factorization (2) = Note that this factorization method assumes assumes that the constants are method that the constants are positive, but that's that’s okay okay since all the the answer answer positive, but since all choices It’s possible that aa = z -‐33 choices are are positive. positive. It's possible that and b = : 2 or a = = 3 and and and b = -‐ 22,, for instance, instance, but but these these are are cases cases that that you you generally generally don't don't need need to worry worry about about for this this factorization. factorization .

[QjFor an equation

to have no solutions, the coefficients terms must m u s t be be the same coefficients of the the x terms the same on either side but but the constants must must be either side the constants different. First, First, let's let’s expand left side. side. different. expand the the left

3 x+ +3 a x= = 1122-‐ 77xx 3x 3aa-‐ 22ax 3 a-‐ 22ax a x = 112 2 -‐ 1lOx 0x 3a

Comparing Comparing the the coefficients coefficients of the the x terms, terms, a= 0 , a = 5. Note Note that is N O T equal -‐ 22a = -‐ 110, that a is NOT equal to 4 since o t for the since the the goal goal is n not the constants constants to be the same.. the same 7.

[II Expand

the right side.

(2x + 3)(ax =12x2 15 3)(ax -‐ 5) = 12x2 + bx -‐ 15

If you this factorization factorization and and you didn’t didn't use use this instead expanded expanded the the right right side, side, you you instead would’ve gotten gotten would've 18x2 ‐ 8 = 2a2x2 ‐

1 2 1 x 2

xX--‐ 5X x -‐ 44 = ‐ 44 -‐' 5xx

2ax2 10x -‐ 15 = 12x2 2ax 2 + 3ax -‐ lOx 12x2 + bx bx -‐ 15 2ax2 0 ) x -‐ 15 1 5= b x‐- 15 15 2ax 2 + (311 (3a -‐ 1lO)x = 1212 12x2 + bx

Comparing 12 and Comparing both both sides, sides, 2a 2a = 12 and b= which yields and = 3a 3a ‐- 10, which yields a =z 6 and b = 33aa -‐ 1100 =: 33(6) ( 6 )-‐ 1100=: 88..

2172

Comparing the either side, side, Comparing the coefficients coefficients on either 18 = 2a2 18 2a2 and and 8 = 2b2. 2b2 . Solving Solving these these equations equations gives $33 and = ± :l:2. you can can see, see, we we gives a a := ± and b = 2. As As you

293

CHAPTER 30 ANSWERS CHAPTER ANSWERS TO THE THE EXERCISES EXERCISES

8. j 49 IExpand Expand the side: the left side:

13. E] @JMultiply Multiply both both sides sides by by xyp. xyp.

x2+6xy+9y2=x2+9y2+42 x2 + 6xy + 9y 2 = x2 + 9y 2 + 42

_1 + 1_ : _1 -+-=xx

/xl+6xy+9f=xz+9 + 6xy + JI = / + JI +42 + 42

w+w=w

x2y2 x2y2= = 49

yp = : X __1E_ p yy -‐ p

9. [fil For an 9. an equation equation to to have have infinitely infinitely many many solutions, both both sides sides must solutions, must beequivalent. be equivalent. First, let’s let's expand First, expand the right side the right side of the the equation: equation:

14.

LE] @]Expand Expand the the left side side of the the equation. equation. (x3 + kx2 - 3) (x - 2) (x3+kx2‐3)(x‐2)

6xx =: xx-‐ 66nx 6 nx+ +33xx 6x 6x = 4x -‐ 6nx 6nx

= x 4 + kx3 - 3x - 2x3 - 2kx2 + 6 =x4+kx3‐3x‐2x3‐2kx2+6 = x 4 + (k - 2)x 3 - 2kx 2 - 3x + 6 =x4+(k‐2)x3‐2kx2‐3x+6

2x == -‐6nx 2x 6nx

Comparing this this to x4 Comparing 18x22 ‐- 3x x 4 + 7x3 7x3 -‐ 18x 3x + + 6, we can see that k - 2 = 77and and ‐- 22kk = wecanseethatk‐ = -‐ 118. 8 . lInn both cases, cases, k = both : 9.

Now when we compare compare the N o w when the coefficients coefficients on both sides, we get 2 = both sides, = -‐ 66n, n , which which gives gives 1 2 n:_‐6= n =- 6 =-3·3.

[j

15. ~ Multiply Multiply both both sides sides by by (x + + 3)(x 3)(x ‐- 2). We We get

Multiply both both sides sides by Multiply by b. b.

5(x 5 ( x-- 22)) ‐- 22(x ( x + 33)) = aaxx -‐ b 5 2 x- ‐ 66 = z aaxx -‐ b 5xx-‐ 1 100- ‐ 2x 3x -‐ 16 ax -‐ b 3x 16 = ax

a b abb++aa = aa++5Sb a b = 5 b ab Sb a=5 S 11.

[IjMultiply Multiply both sides by by x(x x(x -‐ both sides

Comparing the the coefficients coefficients on Comparing on either either side, side, a := 3andb = 16. Therefore, Therefore, 3 and b = a + b ==33++116 6 ==119. 9. a+b

4). 4).

((xx-‐ 4 x ( xx -‐ 44)) 4)) -‐ xx ==x(

lNotice ( x -‐ INotice that that x22‐- 11==( x(x++1 l))(x

1 16. j ;

-‐44 z=x2 x ‐- 4x 4x 2

o (x -‐ 2) 2)22 0= = (x

4 ( x++ l1)+2(x) + 2 ( x ‐ 1) 1 )=: 335 5 4(x 4 5 4xx++ 44 ++ 2 2xx-‐ 22 ==335 6 5 6xx++ 22 z=335 6 6xx = 3333

We can can see that that x = 2.

0

1 1))o onn

right hand hand side. side. It’s the right It's then then easy easy to see see that that we should should multiply we sides by multiply both both sides by ((xx + l ))(x ( x -‐ l1). ).

0 z x2 0= x 2 -‐ 4x 4x + + 44

12.

p p

YP + xp = xy - xxpp yypp = xxyy ~ yp = x(y y p = X ( l / -~ pp))

6xy = 42 6xy = xy xy = 77

10.

yy

Expanding side, Expanding the right right side,

4x2+mx+9 z4x2+4nx+n2 4x2 + mx + 9 = 4x2 + 4nx + n2

xV2 = _ 22 11

Comparing Comparing both both sides, we see see that sides, we that = 4n = 1n122 and and m m = 4n

9 9:

Therefore, n = -‐33 and and m = Therefore, = 4(‐3) 2 4( -3) = -‐ 112

m +n = : ‐ -1 12 2 ++((‐- 33)) = ‐- 115 5 m+

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17. [[) Expand hand side. Expand the the left left hand side.

(2x -‐ b)( 7x + 14x 2 -‐ ex b)(7x + b) = =14x2 cx -‐ 16 16 2 -‐ ex 14x b2 = 14x22 + + 2bx Zinc -‐ 7bx - b2 = 14x 143:2 cx -‐ 16

b2 = 14x 14x 14x22 -‐ Sbx 5bx -‐ b2 14x22 -‐ ex cx -‐ 16

Comparing both can see Comparing both sides, sides, we we can see that that b = =4 (b cannot 0). cannot be be -‐44 because because b > O). 2 Sb 5b = 5(4) = = 20. ec = 18.

@] Multiply both by (n -‐ 1l))(( n + lMultiply both sides sides by(n + 1). 1). 3(n 2n (n -‐ 1) (11+ l )( n -‐ 1) 3(n + + 1) 1) + +2n(n 1) = 33(n +1)(n 1) 3n + 3 + 2n 2 - 2n = 3(112 - 1) 3n+3+2n2‐2n =3(n2‐1) 2 2n2+n+3=3n2‐3 2n + n + 3 = 3n 2 - 3

2 00 = - ‐11 = nn2 n- ‐66

00 := ((n +22)) n ‐- 33)( )(n+

n = 33 oorr -‐ 22.. BBecause e c a u s ne n>>0, 0 ,nn== 33..

295 295

CHAPTER 30 ANSWERS CHAPTER ANSWERS TO TO THE THE EXERCISES EXERCISES

Chapter 10: Systemsof Chapter Systems of Equations Equations CHAPTER EXERCISE: CHAPTER

1. [fil Substituting 1. Substituting the the the second second equation equation into into the

;

6. 6.

first, first,

w o lines [I] IfIf the the ttwo lines intersect intersect at at the the point point (2, (2, 8), 8),

then solution to system. then (2,8) (2,8 ) is a a solution to the the system. Plugging the Plugging the point point into into the the equation equation of of the the second line, second line, we can can solve solve for for b, b,

3 (1 - 3y) - 5y =- ‐ 111 3(1~3y)‐5y= 1 3 -‐ 99yy -‐ 5Syy = z ‐- 111 1 3 -‐ 114y 3 4 y== ‐- 1111

yy = z ‘- bbx" 8 == ‐b(2) - b(2) 8 ‐- 4 = =b

‐- 114y 4 y == ‐- 1144

l yy = =1

Plugging the Plugging the point point into into the the equation equation of of the the

Finally Finally,, x = z 11 -‐ 3(1) = = -‐ 22..

line, first line,

[QJFrom 2. E] From the the first equation, = 20 equation, 31 y= 20 -‐ 2x. 2x. Plugging this Plugging this into into the the second second equation, equation,

ax + +b b yy = : ax 8 = a(2) a (2) -‐ 4

6xx -‐ 55(20 6 ( 2 0-‐ 22x) x ) = 112 2

1 2=: 22a“ 12 6 6 =: aa

6xx -‐ 1100 lOx 6 0 0+ +1 0 x=: 112 2 16x : 112 16x = x = z 7

7.

The two graphs do not intersect at all, so

there are there are no no solutions. solutions.

We already know Wealready know the at this this the answer answer is is (D) (D) at point , but point, z 20 ‐ 2(7) = 6. but just just in case, case, y = 20 - 2(7) = 6. 3.

0

8.

[1]From From the the first first equation, equation , we we can can isolate isolate yy to to get y = -‐ 5Sxx ‐- 2. Substituting get Substituting this this into into the the second equation, second equation ,

[1]Add Add the w o equations the ttwo equations to to get get 7x -‐ 7y 7x 7y = 35. Dividing both sides sides by by 7, Dividing both 7, 5. We can x ‐- y == 5.We can multiply multiply both both sides sides by by ~1 - 1 to get y x = 5. togety‐x=‐5.

2(2x = 33 -‐ 3(‐5x 2(2x -‐ 1) 1) = 3(- Sx -‐ 2) 2)

4 3 3+ + 15 x ++66 4xx-‐ 22 := 15x

4. ~ The The fastest fastest way way to do do this to this problem problem is is to subtract the the second subtract second equation equation from from the the first, first, which yields x + +y = z 9. 9. which yields

4x ‐- 2 := 15x + +9 11x -‐ 1ll1 = = ll x _1 z x - 1=X

[I] In the the first equation, we can m o v e 3x 3x to to equation , we can move

Finally, _5( _1) _ 2 = 3. Finally,yy z = - 5( - 1) - 2 = 3.

5.

the right right hand the : ‐- 5Sxx + + 8. 8. hand side side to to get get y = Substituting this Substituting into the the second second equation, equation, this into 2((-‐ 55xx + + 8) 8) = ”- 33xx + +2 = z -‐ 33xx ‐- 10x lOx + + 16 16 = : -~13x 13x = x Z x=

9_ 9. [1]Divide Divide the the first first equation equation by by 2 2 to to get get x ‐- 2y := 4. We can’t can't get get the the coefficients coefficients to to match match ((-‐ 2 vs. vs . 2 2 for for the the y’s). y's). Therefore, Therefore, the the system solution.. In system has has one one solution In fact, fact, we we can can even even solve w o equations solve this this system system by by adding adding the the ttwo equations to get y= = O. get 2x = 8, 8, x = 4, 4, which which makes makes y 0.

-‘ 110 0 ‐- 110 0 6 -‐ 226 2 2

10.

Then, y = ‐5(2) Then,y - 5 (2) + 8 z = ‐- 22.. Finally, Finally, xy = (2)( 2) = 4. xy = (‐2) ‐ 4 .

the [I] To get get the the same same coefficients, coefficients, multiply multiply the

first equation 10y = equation by ‐2 - 2 to get get ‐- 44xx + + lOy = -‐2a. 2a. Now N o w we a = ‐- 88,, a we can can see see that that ‐- 22a a= = 4. 4.

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THE COLLEGE PANDA THE COLLEGE PANDA

11.

0

the 17. E] ~ Plugging Plugging the the first first equation equation into into the second, second,

Multiply equation by Multiply the the first first equation by -3 ‐3 to to get get -‐3ax 3ax -‐ 6y = 15. The = -‐15. The constant constant a cannot cannot be be -‐ 1. equation 's 1. Otherwise, Otherwise, the the second second equation’s coefficients equal to the coefficients would would then then be be equal the first equation's coefficients, equation’s coefficients, resulting resulting in aa system system

v'4x m ‐ -( fi(v1x ++ 3 3)) == 33

z fi -‐ fiVX ‐ -33== 33 2/x /x fi ==6 6 x z=336 6 X

with with no no solution solution.. 12. @ First, First, multiply multiply the the fust first equation equation by by 33 to to get rid of the fraction: 12x y = 24. Next, get rid the fraction: ‐ = ‐24. Next, substitute substitute the the second second equation equation into into the the fust, first,

Therefore, \/3_6 + +3= : 9. Therefore, y = v'36

12x -‐ (4x (4x + 16) 16) = = -‐ 224 4 18.

12x 1 2 x- ‐4x 4 x- ‐ 16 1 6=: -‐ 224 4

medium, and and large large jars, respectively. Based Based medium, jars, respectively. on the information, we can create on the information, we can create the the following following ttwo w o equations equations::

Bx 8 x = -‐ 8 x X

== -‐ 1

Finally,y 4(‐1) + 116 6= = 12. y = 4(1) + Finally, 13.

16s=2m+1 = 2m + I 16s

4 s +m m = 1I 4s+

[IQ]We Wecan can isolate isolate x in the the second second equation equation to to Substituting this get get x = = y -‐ 18. Substituting this into into the the first first equation, equation,

the weight weight of the the large terms of To get get the large jar jar in terms the weight of the small jar, we need get rid the weight the small we need to to get rid the weight weight of the the medium of m, m, the medium jar jar.. We Wecould could certainly use elimination, but here, we'll certainly use elimination, but here, we’ll use use substitution. Isolating the second substitution. Isolating m m in the second into equation, m equation, m = I -‐ 4s. Substituting Substituting this this into the first first equation, equation, we we get the get

y3]= = O.S (y - 18) + 0.5(y‐18) + 114 4

. 5 y-‐ 99 ++114 4 yy ==00.Sy O.5y=5 0.Sy =5

0 y ==110 14.

16s=2(1~4s)+l 16s = 2(/ - 4s) + /

Cf]To match coefficients, multiply match the the coefficients, multiply the the

16S=21‐SS+1 16s = 21- 8s + I

equation by by 18 18to get 6x 6x -‐ 3y = : 72. We We first equation to get can then can then see see that that aa = = 33 if if the the system system is to have have no solution solution.. no

24s=3l 24s = 31 8 8s5=: 1I

E’ Divide Divide the the fust first equation equation by by 3 to to get get 15. [Q] 2y = 5. Divide Divide the the second second equation equation by by -‐22 x -‐ 2y to get get x -‐ 2y 2y = : 5. They're They're the the same, same, so so there there are are an an infinite infinite number number of solutions solutions.. 16.

Eight small jars jars are needed to match Eight smaJl are needed match the the weight large jar. weight of one one large E] Since were 30 30 questions, questions, James James must Since there there were must 19. [Q] have z 30. The The points points have had had 30 answers, answers, x + y = he 5x. The The he earned earned from from correct correct answers answers total total 5x. points he lost lost from answers total points he from incorrect incorrect answers total 2y. Therefore, 5x 5x -‐ 2y 231 = 59. Therefore,

For aa system system to to have have infinitely infinitely many many Cf]For

solutions, solutions, the the equations equations must must essentially essentially be be the same. same. Looking Looking at the the constants, constants, we can can the make them them match match by by multiplying multiplying the the second second make equation by 2. The The equations equations then then look look like like equation this:: this

Leta and points you you get get be the the number number points 20. @] Leta and b be for hitting hitting regions A and and B,respectively. regions A B, respectively. From information, we form the the From the the information, we can can form following t w o equations: following two equations :

mx -‐ 6y 6y =: 10 mx 10 4x -‐ 2ny = 10 10 4x

+ 22bb = 118 8 a+ 2 a + b = 2 2a + b 211

N o w it's it’s easy easy to to see see that that m m= z 4 and and 2n Zn = : 6, 6, Now

.

m_ 4

n =‐ 3. 3. Finally, Finally, ;: n

[II Lets, Lets, m, and I be be the the weights ll, m, and weights of sma small,

‐ ;.5. =

To solve solve for b, b,multiply first equation equation by by 22 multiply the the first and subtract to get get 3b = 15, 15, b = 5. 5. and subtract 3b =

297

CHAPTER 30 ANSWERS TO THE EXERCISES

21.

[Q]Let r and

c be the number of rectangular tables tables and and circular circular tables, tables, respectively, respectively, at at the the restaurant. Based on on the restaurant. Based the information, information, we we can can make make the the following following two t w o equations: equations:

24.

II] To find the point(s)

where two graphs intersect, solve the the system system consisting their intersect, solve consisting of their equations . In this this problem, problem, that that system system is equations.

y=x2‐7x+7 y = x 2 - 7x + 7 : 22xx -‐ 1 y=

4r+ + 88cc ==1144 44 4r rr

+ +cC ==3300

Substituting the first equation equation into the Substituting the into the second, we get second, get

To solve solve for r, multiply multiply the the second second equation equation by 8 and and subtract subtract to get get -‐ 44rr = : -‐ 996, 6 , r = 24.

2 - 7x + 7 = 2x - 1 xx2‐7x+7=2x‐1

22. ~ The The solution solution to the the system system is is the the intersection point of the two lines. intersection point the t w o lines. Each Each

2 - 9x + 8 = 0 xx2~9x+8:0

(x( x ‐ 1l )(x) ( x ‐ 8)= 8 ) = 00

horizontal horizontal step step along along the the grid grid represents represents ~2 of

xl e= o1 or r 8

aa unit, and each unit, and each vertical vertical step step along along the the grid grid represents represents 1 unit. unit. So, So, the the intersection intersection point point is 3 at( - ~, - 3).

So the the x-coordinates x-coordinates of the the points So points of intersection are and 8. Since intersection are 1 and Since the the question question must be already gave (1, 1), p must already gave us us the the point point (1,1), be

“(74>

23.

equal to 8. equal

[Q]From the second

equation, x = 2y. Plugging the first equation, Plugging this this into into the equation, we we get get

I

25. 9 or 16 j First, both sides sides of the First, add add 11 to both the second equation to second equation to get get y =: x + 11. Then Then substitu te this this in for y in the the first equation: substitute equation:

2 _ 2= : _1_ (2y)2 (231) _ y2 y 12 12 2 _ 2 == -1_ 4y2 4y - y2 y 12 12 1_ ‐ _1 3y22 = 3 y ' i-12z

x 2 - 2x = x + 11 - 1 x2‐2x:x+11‐1 2 - 3x - 10 = 0 xx2‐3x-1ozo ((xx ++ 2 ) ( x- ‐ 5) 5 )= =0 2)(x

=-‐ 22or5 x= 0r5

1

y2=l = 36 2

y

When +1111 = 99.. W When W h e nxx== -‐ 22,,yy == -‐ 22 + h e nxx == 55,, The so lutions to the the system y := 5 + 11 = 16. The solutions system are then then (‐2,9) (- 2, 9) and and (5,16). (5, 16). Therefore, the are Therefore, the possible values possible values of y are are 9 9 and and 16.

1

yy = ‐ i±-g 6

1 Therefore, Therefore, the the values values of y1 and and y2 y; are are -‐2 and and

6

1

6. 6

298

THE COLLEGE COLLEGE PANDA THE PANDA

Chapter11: Chapter 11: Inequalities Inequalities CHAPTE R EXERCISE :

1.‑ -‐ xx ‐- 44>>44x x ‐- 114 4

‐- 55xx > ‐-110 0 x< §x‐10 x - 4 > - x - 10 4 2

3xx-‐ 1166 > 22xx -‐ 4400 3 xX >> ‐-224 4 3. 3.

The shaded shaded region below the the horizontal : 3, above [g The region falls below horizontal line line 31 y= 3, soy soy < 3. 3. The The shaded shaded region region also also stays stays above

y= soy = x, X, SO y > x. X. 4.

Let’s say say Jerry's Jerry's estimate, estimate, m, m, is 100 marbles. marbles. [fil Let's

5. 5.

[I] Setting Setting up up the the inequality, inequality,

If the the actual actual number of marb marbles is within within 10 If number of les is 10 of of that that estimate, estimate, then then the the actual number must m u s t be be at at least least 90 m ‐- 10 5 n S actual number 90 and and at at most most 110. Using Using variables, variables, m 10 ::; 11 ::; + 10. m+ lO.

M 2N N M ?. 12P + +100 12P 100 2 ?. ‐- 33PP -++ 970 970 15P 2 15P ?. 870

P ?. 2 58 58 6.

[zJ 3 ( n-‐ 22)) > -‐ 44(11 ( n -‐ 99)) 3(11

3 n-‐ 66 >>‐ -44n n ++336 6 3n 7 2 7nn > 442 n>6 Since n11is Since is an integer, the least possible value of of 11 n is 7. an integer, the least possib le value is 7.

7.

The shaded shaded region region is is below below the the horizontal horizontal line line y3;= ‐ 3 . Therefore, Therefore, [fil The = 33 but but above above the the horizontal horizontal line line y = = -3.

y2 ?. ‐3 - 3and and y 5 ::;3. 3.

8. 8.

16 hours. Since spends on the bus bus is 2 hours he spends spends on the train train is is % [I] The The time time Harry Harry spends on the is ~ hours and and the the time time he on the hours. Since

y

X

16 8 16 the the total total number number of hours hours is never never greater greater than than 1, z 1. - + + y- S::;1. X

299 299

y

CHAPTER 30 ANSWERS TO THE EXERCISES

9.

10.

[I]

U the the distributor distributor contracts If u t to Company u t to for contracts oout Company A A for for xx hours, hours, then then it it contracts contracts oout to Company Company B B for 10 ‐- x hours. 10 Company A then produces 80x cartons and Company B produces 140(10 ‐ x) cartons. hours. Company then produces cartons and Company B produces 140(10 - x ) cartons. Setting up Setting up the the inequality, inequality, 80x + +140(10 1,100 140(10 ‐- x) > 1,100

[El [QJPlug Plug in x = =

l,1, y z= 20 to get 15+ > a. 20 into into the the first first inequality inequality to get 20 20 > > 15 + a, a, 55 > a. Do Do the the same same for for the the second second inequality 15 < b. is less 15. The between inequality to get get 20 20 < 55 + b, b, 15 b. 50, So, a a is less than than 5 5 and and b bisis greater greater than than 15. The difference difference between the wo m u s t be m o r e than = 10. 10. Among 12 is is the only one greater than the ttwo must be more than 15 ‐- 5 = Among the the answer answer choices, choices, 12 the only one that that is is greater than 10.

11. E, [QJThe line = ~x 2x + +2 line going going from from the the bottom-left bottom-left to to the the top-right top -right must must be beyy = 2 and and the the line line going going from from the the top-left to the the bottom-right top-left = -‐ 22xx + + 55 (based bottom-right must must be beyy = (based on on the the slopes slopes and and y‐intercepts). y-intercepts). Answer Answer (D) (D) correctly shades shades in the correctly the region : ~2xx + + 22 and and below region above above y = below yy = ‐- 22xx + + 5. 5. 12.

[I] One One manicure manicure takes takes 1/3 hour. One One pedicure she 1 /3 of an an hour. pedicure takes takes 11//22 an an hour. hour . The The total total number number of of hours hours she

1

1 spends doing doing manicures spends must be be less 30, so so ~m 3m + + 5pp ~ 5 30. manicures and and pedicures pedicures must less than than or or equal equal to to 30, 30. She She earns earns the manicures manicures and 25m for the pedicures. Altogether, Altogether, 25m and 40p for the the pedicures. 25m + + 40}: 40p 2 2: 900. 900.

13. |E [Q]From From the given -+- 12. Subtracting Subtracting 12 12 5 which given inequality, inequality, x 5 ~ 3k 3k + 12 from from both both sides sides gives gives x x ‐- 12 ~ 3k, 3k, which confirms that that I is always confirms always true. true. From the the given given inequality, inequality, 3k + 12 From 2 k, which means 2k 2 2, k u s t also 12 2: which means 2: -‐ 112, k2 2: ‐- 66,, so so 11 II m must also be be true. true. From the the given given inequality, inequality, k 5 From I“ must ~ x. Subtracting Subtracting k from from both both sides sides gives gives 0 OS ~ x x -‐ k. k. Therefore, Therefore, III must also also be true be true.. First, 14.13~ < x < ?~ ILet’s Let's solve solve these these separately. separately. First, 2 20

‐30 < -‐ 22xx ++ 4 -3

0 < -‐ 66xx + 1 -‐ 220 122 ‐- 332 2 -r_ |- x

a :::::28.4 students students Since it wouldn't wouldn’t make make sense sense to have have four -tenths of aa student, four-tenths that can can be student, the the most most that be accommodated is 28 accommodated 28 students. students.

0

To get get the the minimum minimum number number of greeting greeting cards the shop could have cards the shop could sold, we assume have sold, assume that the shop sold sold as as many many gift boxes that the shop as boxes as possible (400 gift boxes). possible boxes). Since each each gift box was sold sold for $7, the store was store sold sold 400 x $7 $2,800 $7 = = $2, 800 worth worth of gift boxes boxes in this scenario, which scenario, least which means means the the store store sold sold at least $8,000 $8, 000 -‐ $2, 800 = = $5, 200 worth $2,800 $5,200 worth of greeting greeting cards . Since each cards. each greeting greeting card card was was sold sold for store could could have $5, the store have sold sold a minimum minimum of $5,200-;- $5 = 1,040 greeting cards to meet its goal. goal.

8. @] Giovanni made 0.15 (25)( 12) = $45 during lunch. lunch. If he during tables during during he serves serves x tables dinner, he he will make dinner, make an additional 0.15(45)x an additional 0.15 (45)x dollars. Since the dollars. should be be the total total for the the day day should least $180, $180, at least

45 + 0.15(45)x 2 180 45 + 0.15(45 )x 2:: 0.15(45)x 2135 0.15 (45 )x ~ 135 x X

1,800 the salon salon needs needs at least 1,550 = : 22.5 ➔ ‐> 23 least ~

toolkits. To get get the required required number toolkits. number of nail nail buffers, the buffers, the salon salon needs needs at at least least 4,000 4 , OOO::::: 27 toolkits. toolkits . Based m 26.7 ➔ ‐‐> 27 Based on these these 150 numbers, a minimum numbers, minimum of 27 27 toolkits toolkits must must be be purchased the salon purchased for the salon to receive the receive both both the required number of nail files and the required number of nail number nail buffers. buffers.

0

Two liters is equivalent to 2 x 33.8 = 67.6 ounces, which will fill 67.6 -;- 12 :::::5.63 plastic cups. cups can be cups. Soat So at most, most , 5 plastic plastic cups can be completely completely filled.

304

135

~ o.15(45) 0.15 (45)

x ~2 X

4. I27 J To get required number get the the required number of nail nail files,

5.

[m

0 20

TI-IE COLLEGE PANDA THE COLLEGE PANDA

9. [!!] Leta Let a be be the the number number of fish fish Ashleigh Ashleigh caught caught and and nn be be the the number number of fish Naomi Naomi caught. Using these caught. Using these variables, variables, we can can set set up up aa system system of an an equation equation and and an an inequality: inequality:

11. []] Because Because we're trying to maximize the we’re trying maximize the number of nighttime assume that that nighttime bottles, bottles, we we assume number 65 daytime bottles bottles were 65 only 65daytime only were filled. filled. The 65 daytime bottles 65 x 2 = 130 ounces daytime bottles used used up up 65 130 ounces the active active ingredient ingredient and and 65 of the 65 x 66 = 390 ounces of flavored flavored syrup, syrup, leaving leaving ounces the active active 385 -‐ 130 130 = = 255 ounces ounces of the ounces of ingredient and ingredient and 850 -‐ 390 = 460 460 ounces flavored syrup ounces of flavored syrup.. The The remaining remaining ounces 255 . ingredient . d'1ent are active mgre are enough enoug hf for or ¥ = active : 85 85

a = 33nn-‐ 9

45 aa + n z2:45 The The equation equation allows allows us us to substitute substitute for for a in the the inequality: inequality:

3

aa+r1245 + n 2:45 (3n - 9) + n 2: 45 (3n‐9)‐+‐n245 4n254 2: 54 4n n213.5 n 2: 13.5

nighttime bottles, bottles, and the remaining ounces nighttime and the remaining ounces 460

flavored syrup enough for g of flavored syrup are are enough

5

= 92 = 92

nighttime bottles. bottles. Based Based on nighttime on these these numbers numbers,, we're limited remaining amount amount of we're limited by the the remaining active ingredient, ingredient, so so the the maximum maximum number number of active nighttime bottles nighttime bottles that that can can be be filled is is 85.

Since Since it's it’s implied implied that that fish are are caught caught in whole whole numbers, numbers, the the minimum minimum possible possible value value of n is 14. 14.

12. [Q] Let's set set up a a system as the the El Let’s system withs with s as as the the number number of number of short tables and and I as number short tables long tables. tables. long

10. I125 IIf If we we let let b be be the the number number of black black pebbles, pebbles, w w be be the the number number of white white pebbles, pebbles, and and j be be the the number number of jade jade pebbles, pebbles, then then

4 S++ 8/ 8 l ==1168 68 4s

j > ~ g and and w < 2b. Since j = 32, 32, the the first

s + Il §::;3322

inequality inequality becomes becomes 32 32 > ~, g, which which simplifies simplifies

Divide both sides sides of the the equation equation by 4 to get Divide both get

to 64 so the 64 > b, b,so the maximum maximum possible possible value value of bb is 63. Using Using b = 63, the the second second inequality inequality becomes becomes w w < 2(63), which which simplifies simplifies to w < < 126. Based Based on this this result, result, the the maximum maximum possible possible value value of w w is 125. 125. Now N o w you you might might be be wondering why we wondering why we used used the the maximum maximum in the second possible possible value value of bbin second inequality. inequality. Since Since w w is less less than than 2b, maximizing maximizing w w means means that we have to maximize that we have maximize b first.

21:= 42. 42. Isolating Isolating I then 21 -‐ ~. 2. s5+ + 21 then gives gives /1= = 21 Substituting this result into into the inequality, we we Substituting this result the inequality, get get s s + 21 - -2 < 32 -

:. < 11 2 -

s::; 22 Based on this result, the maximum number of Based on this result, the maximum number short used is 22. short tables tables that that can can be be used

get at least $140 worth worth of tacos, tacos, a 13. [!!] To To get least $140 customer would have have to receive customer would receive at least least 140 53.8 ➔ ‐> 54 tacos tacos (we (we round round up since since ~m 53.8

~-!~ m

it’s implied tacos are are given given in whole it's implied that that tacos whole numbers only). To receive 54 tacos, numbers only). receive at at least least 54 tacos, the would have have to buy buy at least the customer customer would least 54 = 13.5 ➔ ‐> 14 14burritos (again, we round up up = burritos (again, we round 4 since implied that are sold sold in since it’s it's implied that burritos burritos are whole numbers). Therefore, Therefore, 14 the 14 is the whole numbers). minimum.. minimum 54‐4

305

CHAPTER 30 ANSWERS TO THE EXERCISES

14.

[I] Let a be the number

the numbers 16. I54 ILet s, m, and and Il represent represent the numbers of

of two-tier cakes and be the number b be number of three-tier cakes Ava three-tier cakes decorates . Using decorates. Using these variables, we we can can set set these variables, up the the following following system up system of inequalities: inequalities:

small, medium, medium, and small, boxes shipped, and large large boxes shipped, respectively. Based respectively. Based on on the the information information given, given,

50 ss + m m + II==2 250

2 0 a+ +3 5 b~ § 3360 60 20a 35b

s+m lI > >s+

a+b214 a + b 2'.14

Since m = = 70, the the system system becomes becomes

Note converted 6 hours 360 Note that that we converted hours to 360 minutes to set set up minutes inequality.. The up the the first inequality The first inequality 4a+ 7b 5 inequality then then simplifies simplifies to 4a + 7b ~ 72. To solve this solve this system, system, we have get the the signs signs have to get pointing in the pointing the same same direction so that we can direction so that we can add the add Remember that the inequalities. inequalities. Remember that inequalities can inequalities can be be added added only only if if their their signs signs point in the the same same direction point (do not not subtract subtract direction (do inequalities; think inequalities; only in terms terms of adding adding think only them) . Soif So if we we multiply multiply the them). the second second inequality by -‐ 44,, we inequality up with the we end end up with the following system system (note following (note the the sign sign change): change) :

80 ss + II ==1 180 0 Il >> ss ++770

IsolatingI Isolating l in the the equation equation gives gives Il = = 180 ‐- 5. s. Substituting this Substituting inequality, we we get get this into into the the inequality, 1 8 0- ‐ss > ss ++7700 180 110>25 110 > 2s 5 >s 555 > Since Since s5is implied to bea whole number, number, the the is implied to be a whole greatest possible greatest possible value value of s5 based on this based on this result is 54. result

4a+7b 4a + 7b S ~ 72 72 -‐ 44aa -‐ 44b bS 6 ~ -‐ 556

17.

Adding the Adding 16, the inequalities, inequalities, we we get get 3b 3b 5 ~ 16, which simplifies which 5 5.33. Since Since it's simplifies to b ~ it's implied implied that that Ava decorates whole decorates cakes cakes in whole numbers, numbers, the the maximum maximum possible possible value value of b is 5. This question could’ve also been solved solved question could've also been through guess through guess and and check. check.

[zJBased Based on on the the given 11= 10 given information, information, n = 10 and w = and = Bx 8x (8 ounces ounces of water water in each each of the the cups) . To use use these x cups). these values, we first set set up values, we the following the inequality: following inequality: CS ~ 16% 16%

100n < 100n < 116 6 nn + w -

15. [RJ E] For For 11 pound seasoning, Lianne Lianne will will pound of seasoning, need 0.75 pounds need salt and pounds pounds of sea salt and 0.25 pounds black pepper. of black The sea salt pepper. The salt will will cost cost 0.75 x $2 the black pepper $2 = $1.50 and and the black pepper will will cost 0.25 x $8 Altogether, that' that’ss $8 = $2. Altogether, $1.50 + + $2 = = $3.50 for each each pound pound of seasoning . Since Lianne seasoning. can spend more Lianne can spend no more than $210, she than she is limited limited to making making . $210 : 60 Therefore, $ = 60 pounds pounds of seasoning. seasoning . Therefore, 350 ssfso 60 maximum. 60 is the the maximum.

100(10) 10 < 16 100 10 + 8x s- 16 i 8.3 1000 ~ S 16 16(10 8x) 1000 (10 + Bx)

1000 ~ S 160 160 + + 128x 128x 1000 840 S ~ 128x

6.56 5 ~ x X Since the question Since question indicates indicates that that x is aa whole whole number, number, the the minimum minimum possible possible value value of of xx based to note note based on on this this result result is 7. It’s It's important important to that we were thatwe sides by by were able able to to multiply multiply both both sides 10+ sign 10 + 8x Bx without without worrying worrying about about a sign change because 10 10+ 8x is guaranteed be change because + Bx guaranteed to to be positive (x must positive must be be positive positive in the the context context of the problem). problem) . the

306

THE COLLEGE COLLEGE PANDA THE PANDA

Chapter 14: 14: Lines CHAPTER EXERCISE EXERCISE::

1. 1.

[filA vertical line that vertical line that intersects intersects the the x-axis at

5.

3 has has an an equation equation of xx = z 3.

2.‑ 2. [f]

of line/ goes up three units for every ttwo every w o units its units to the the right, right , which which means means its

slope is ~. 2 A parallel line must must have have the slope parallel line the same same slope. Only slope. Only answer answer choice choice (C) (C) gives gives an an equation slope. equation of a a line line with with the the same same slope.

- y1 z}1 yz‐y1

Y2

xX2z -‐ xX)1 n l 7 1- ‐ 1

3 1

1)

=3

n -‐ 1l --=

1 -

5 -(ng

6.

B

From the graph, we can see that

J

goes up to the up 11 unit unit for every every 2 units units to the right, right,

6 =3 6 3

1 which which means means its slope slope is ~.. Since g is

E

n -‐ 11 = 2

perpendicular to J, f, the the slope slope of gg must must be perpendicular be -‐22 (the negative reciprocal). g passes (the negative reciprocal). Since g passes

n=3

3. 3.

[f] The graph

through point (1, we can through the the point ( 1, 2), ~), we can use use

0

Draw the x-intercept to Draw aa line line from the x-intercept of -‐22 to the y‐intercept y-intercept of -‐ 44..

point-slope equation of point -slope form to find the equation of g: yy ‐- yy1 1 ==" m(x 1 ( x-‐ xxi) i) 5 y‐§‐‐2(x‐1) y --=2(x - 1)

yll

2

5 2x+2+2 yy == _- 2x+ 2 + -_ 2

9 y‐‐2x+§ y = - 2x + 2

.

‐4

Finally,g (- 1) = - 2(- 1) +

229 = 2 + 4~2l =

l

6! 2·

A quicker been to work quicker way would've would've been work

you can see, it goes units down down for every every As you can see, goes 4 units 4 2 units to the right. The slope slope is is _74 units to right. The ~ = -‐ 22..

backwards backwards from from the the point point (1, ( 1, g), ~), knowing knowing that the slope slope is -‐ 22.. So Soon graph of g, 11 that the on the the graph of g, unit brings us us to unit to the left brings

. . 8 ‐ 5 ‐_ 9 ‐_ 4.-Theslopeof1mells 4. 0 The slope of line / is 6‐‐(‐_3) ( ) = ~ = !3.. 9 6- - 3 3

(0,2. = (0,4%>, ( 0, ~ + + 2) 2) = (0,41), and and 11 more more unit unit to to the the

Using Using point-slope point‐slope form, form,

left brings brings usto +2) = ((‐1,6%). us to ((‐1,4% - 1,41 + 2) = - 1,6D.

=

m(x x1)) yy -‐ yY1 l =m ( x - 11 1 yy‐8‐§(x~6) - 8 = (x - 6)

Therefore, g( g(‐1) Therefore, - 1) = 6.5.

3

7. []]

1 yy = éxx + +66

3

u = fl2 -=0 =i 2_ = _ 1l xz‐xl 0 -‐ 4 5 -‐ 44 2

answer choice choice by At this this point, point, we we test each each answer plugging x-coordinate and and verifying verifying plugging in the x-coordinate the y‐coordinate. Only answer (A) ( A ) works. they-coordinate. Only answer works.

8.

307

[filFrom the graph,

slope mis positive and y-intercept mb < 0. y-intercept b is negative. negative . Therefore, Therefore, mb 0.

CHAPTER 30 ANSWERS TO THE EXERCISES

@The = -‐ 22xx -‐ 2 9. [Q] The line line y = 2 has has a a slope slope of of -‐22 and a a y‐intercept and u s t have have a y-intercept of -‐ 22.. Line Line Il m must a slope reciprocal of of -‐ 22,, slope that that is the the negative negative reciprocal 1 which which is 2'. Since Since they they have have the the same same

from from the the table. table . Answer Answer A A works works for for Monday Monday (c(7.2) = = 30(7.2) + 400 = = 616) but o t for any 30(7.2) + but n not for any of the days. Answer B works for the other other days. Answer works for Saturday Saturday (c(8.5) (c(8.5) := 60(8.5) 60(8.5) -++ 210 210 = = 720) 720) but but n o t for any not any of the the other other days. days. Answer Answer D does does not value of c for not give give the the correct correct value for any any of the the given given values values of 5. s. Only Only answer answer C C gives gives the the correct value correct value of c for each each of the the given given values values s. These These types of 5. types of questions questions require require you you to be thorough. thorough . Don’t be just test and Don't just test one one case case and choose the choose the first thing thing that that ”works.” "works." You have have evaluate all all the the answer to evaluate answer choices. choices.

2

y-intercept, of line must be be y-intercept, the the equation equation of line Il must yyzix‐Z. = 1 x - 2.

2

10. []] The line we're looking for must have a 1 slope that that is slope is the the negative negative reciprocal reciprocal of of %, ,

2

which is ‐- 22.. which

13. [E [Q] A line line with with a a positive ot positive y‐intercept y-intercept will will nnot cross the the y-axis y-axis at at a negative negative point. cross point. Therefore, Therefore , when x is 0, y cannot cannot be when negative, which which be negative, makes (E) makes answer. (E) the the answer.

y = ‐2x +b - 2x+b Plugging Plugging in the the point (1,5), point (1,5 ),

~

5 =: ‐2(1)+ - 2(1) + bb

1.6 or 2 IFirst, First, plug the point 14. 1l.6 plug the point (2,6) (2, 6) into into the the

7=b

equation of the the line line so so that that we equation we can can solve solve for a: a:

Now that that we we have have b, the Now + 7. the line line is y = -‐ 22xx + 7.

1 a(2) a(2)‐§(6)=8 - (6) = 8

3

@

11. [Q]

2 a- ‐ 22 =z 8 2a

10 -4 _2 2 10‐4 = 3x ‐- l 1 _ 3

2 0 2aa =110

a=5

6 _z 2 6 x -1 ‐ l ‐ 33 X

So the the equation equation of the So the line line is 5x -‐

Cross multiplying, Cross multiplying,

which which gives gives x = g ~ == 1.6.

2 x-‐ 22 ==118 8 2x 2 0 2xx = 220

12.

8. 8.

The x-intercept x-intercept always The y-coordinate of of always has has aa y-coordinate 0, so if we we plug 0,soif we get get 5x = 8, plug in Ofor y, we 5x = 8,

2 ( x-‐ 11)) = 118 8 2(x

x X

1 all z 3y =

15.

==110 0

[1JOne One easy easy way way to approach approach this this problem problem is is make up up numbers to make numbers for a and and b. b. Let Let a = 1 and b = 2 so and that ~g = ;%.. Since so that Since the the second second line line

[SJWe can can use t w o days days use the the values values from from any any two find the the line line that best models to find that best Let’s models the the data. data. Let's use the the values values from from Monday use Monday and and Thursday Thursday to 616 -‐ 584 . calculate the the slope: slope: ---calculate = 80. At this this = 7.2 -‐ 6.8 point, we can tell point, we can tell the the answer answer is probably probably going to be be choice choice C, but going find the but let’s let's find the y-intercept just y-intercept just to be be sure. sure. Currently, we have have Currently, we 80s + c= z 805 -+- b. Plugging = 7.2 and = 616 Plugging in inss = and c c= from Monday, Monday, we from : 80(7.2) ‐+‐ b, we get get 616 = 80(7.2) + which gives gives 1)z which b = 616 ‐- 80(7.2) 80(7.2) = 40. Therefore, c(s) c(s) = 80s 805 + + 40. Therefore,

d perpendicular to the is perpendicular the first, first, C;= - = -‐ 22,, which which e

satisfies the the condition satisfies condition in answer answer choice choice (A).

An alternative alternative solution solution is to test test each each of the the answer choices choices by answer in the the values values by plugging plugging in

308

THE COLLEGE PANDA THE COLLEGE PANDA

Chapter15:Interpreting LinearModels CHAPTEREXERCISE: 1. [I] means the decreases by 33 feet each each day. E The The slope slope is -‐ 33,, which which means the water water level level decreases

E The value value 18 18refers refers to the the slope slope of -‐ 18, which which means loaves remaining 18 means the number number of loaves remaining decreases decreases by 18 2. []] each bakery sells loaves each each hour. hour . each hour. hour. This implies implies that that the the bakery sells 18 18loaves 3.

[g They -intercept of 500 means that when n = 0 (when there were no videos on the site), there were 500 members. members.

4. @ The number 2 refers to the slope of - 2, which means two fewer teaspoons of sugar should be added (C) and which for every Don 't be fooled by answers answers (C) every teaspoon teaspoon of honey honey already already in the the beverage. beverage. Don’t be fooled and (D), which (hands, slope is always always the change in y for each "reverse" ”reverse” the x and and they the y (I: and s, in this this case). case). The slope the change each unit unit increase increase in x, not n o t the other other way way around. around. 5.

[I]

The salesperson earns a commission, but on what? The amount of money he or she brings in. To get that average price that,, we we must m u s t multiply multiply the the number number of cars cars sold sold by by the average price of of each each car. Since xx and and c6 already already represent cars sold, respectively, the represent the commission commission rate rate and and the the number number of cars sold, respectively, the number number 2,000 must must represent represent the the average average price price of each each car.

6. [) The number 2,000 refers to the slope, which means a town 's estimated population increases by 2,000 for each each additional additional school school in the the town town.. 7.

[I]

8.

[fil When t =

9.

[I]

The number 4 refers to the slope of - 4, which means an increase of 1° C decreases the number of words, the the milk hours faster. hours hours until until aa gallon gallon of milk milk goes goes sour sour by 4. In other other words, milk goes goes sour sour 4 hours 0, ther e is no time left in the auction. The auction has finished. Therefore, the 900 is the auction price price of the the lamp. lamp. final auction

Because it's the slope, the 1.30 can be thought of as the exchange rate, converting U.S. dollars into euros. But after after the conversion, 1.50 is subtracted subtracted away, which means means you you get get 1.50 euros you euros. away, which euros less than than you the conversion, should have have.. Therefore, Therefore, the best interpretation interpretation of the y-intercept is a e u r o fee the the bank bank charges charges should the best the 1.50 y-intercept a 1.50 euro to do the conversion. conversion.

. . ~ 22 99 see the answer answer more more clearly, clearly, we we can can put the equation equation mto mtoy = = mx mx+ form: tt == 5xx + 5 . The The slope slope 10. L:.!JTo see put the + bb form:

. To

5

5

2 is ~, 5, or 0.4, which which means means the the load load time time increases increases by 0.4 seconds each image image on the the web web page. page. seconds for each The slope slope is is the the change change in y (daily (daily profit) profit) for each (cakes sold). 11. @ The each unit unit change change in xx (cakes sold).

Notice that the y-intercept negative. It is when no no cakes Therefore, is the the bakery’s bakery 's profit profit when cakes are are sold. sold. Therefore, 12. [) Notice that they -intercept is negative. anything that that varies varies with with the the number number of cakes cakes sold sold is incorrect. example, answer answer (D) is wrong anything incorrect. For example, wrong because because the cost of the cakes cakes that that didn didn't depends on on how many It’s n o t aa fixed number number ' t sell depends how man y the the bakery bakery did did sell. It's not like they the y-intercept The best best interpretation interpretation of the the cost of running running the the bakery bakery (rent (rent,, like -intercept is. The theyy‐intercept -intercept is the labor, machinery,, etc.), which which is likely likely aa fixed number number.. labor, machinery

13.

The solution solution (5,0) (5,0) means means that that the the bakery's bakery's daily zero when when 5 cakes are are sold. sold. [J) The daily profit profit is zero

Therefore,, Therefore

selling five cakes cakes is enough enough to break break-even with daily selling -even with daily expenses. expenses.

slope of the equation equation is 5, which which means means the temperature goes up by 5 degrees degrees every every hour hour.. So 50 the temperature goes up 14. I2.s j The slope every half half hour minutes), the the temperature goes up = 2.5 degrees. every hour (30 minutes), temperature goes up by 0.5 x 5 = degrees.

309 309

CHAPTER 30 ANSWERS TO THE EXERCISES

.

.

.

1 1x

mx

1

15. [E [[I Putting Putting the the equation equation m t o y = mx + + b form, form, y = 5x + one more more turtle into + 7. The The slope slope of % means means that that one turtle requires additional half requires an an additional half aa gallon gallon of water water.. So SoIll 11] is true. true . Getting x in terms terms of y, x = 2y Getting The "slope” water can can support support ttwo wo 2y ‐- 14. The "slope " of 2 2 means means that that 11 more more gallon gallon of of water more turtles. turtles. SoI more So I is true. true. 16.

CgBecause Because this this question question is asking asking for the the change change in "x" ” x ” per y ” (the reverse of slope), we per change change in ”"y" (the reverse of slope), we need need the equation equation to get x in terms to rearrange rearrange the C. terms of C. C = 1.5 + + 2.53: 2.Sx

Dividing each Dividing each element element in the the equation by 2.5, equation by : 0.6 + x 0.4C = = 0.4C -‐ 0.6 x =

The slope slope here here is 0.4, which means the The which means the weight weight of aa shipment shipment increases increases by 0.4 pounds pounds per per dollar dollar increase increase in the the mailing a 10 increase in the mailing cost is equivalent equivalent to a mailing cost. cost. So 50 a 10 dollar dollar increase the mailing a weight weight increase increase of 10 pounds. 10 x 0.4 = : 4 pounds.

310

THE COLLEGE COLLEGE PANDA THE PANDA

Chapter Chapter 16: 16: Functions CHAPTER EXERCISE: CHAPTER EXERCISE:

l.1.

[Q] FE] Check Check each each answer answer choice choice to see see whether whether

10.

J (O) == 20,J f(0) 20,f(1) and /f(3) The only only (l ) = 21, and (3) == 29. The function that that satisfies satisfies all three function three is (D). (D).

2. [Q]J @f(x) t w o graphs graphs (x) := g(x) g (x) when when the two intersect. They intersect intersect. intersect at 3 points, so there points, so there must be be 3 values must values of x where x ) = g(x). where Jf ((x)

conditions . conditions. 11. lg(2) [I)g( 2) =: 22 22 -‐ 11= = 3. 113.So, So, 2+1= ff(g(2)) (g(2)) = J (3) = 3 = f(3) = 32+1 2 110. 0.

(3) = -‐ 22.. N 3. @ Jf(3) o w where ? Now where else else is ffatat -‐ 22? When x = When z ‐- 33.. Soa be -‐ 33.. So a must must be 4.

12. ~ B The The difference difference between 2x22 -‐ 2 and between 2x and 2x + + 4 is a constant In other constant of 6. 1n other words, 6 needs to be needs = 2x 2x22 -‐ 2 to get be added added on to y = 2 y= 2x + : 2x2 + 4. entails a translation translation 6 units units 4. That That entails upward. upward.

[I] Draw Draw aa horizontal at y = 3. 3. This line horizontal line at line intersects intersects J[ ((x) x ) four times, there are four times, so there are four solutions (four values values of x for which solutions (four which

J (x ) = f(X) = 33). )‑ 5.

@ The x‐intercepts x-intercepts of ‐3 - 3 and and 2 mean mean that that fJ((x x ) must and (x -‐ 2). 2). must have have factors of (x + 3) and That A y-intercept y‐intercept of That eliminates eliminates (C) and and (D). A when we plug = 0, 12 means means that that when plug in x = x) = answer (B) meets Jf ((x) = 12. Only Only answer meets all these these

[g Plug Plug in -‐33 and and 3 into into each each of the answer answer

13.

whether you choices to see whether you get get the same the same value. you're smart smart about about it, realize value. If If you’re it, you'll you’ll realize 2 , which always that answer answer (C) has has an an xx2, that always gives gives a positive positive value a value.. Testing Testing (C) (C) out, out, 3(‐3)2 +11 = 28 and Jf ((-‐ 33)) = 3(3)2 + 28 and 2 = 3(3) 3(3)2 + +11 = = 28. The answer indeed ff(3) (3) = answer is indeed

[I) Draw Draw a a horizontal z c, horizontal line line at at y = C, passing passing through (0, c). This horizontal horizontal line through line intersects intersects with f three times. That with three times. That means means there there are are 3 values values of x for which x) = = c. which Jf ((x)

14. 14.@

(C). (C)-

g(k) = 8 8 g( k) =

6. @] First, g(10) = 6.-First,g(10) = f(20) o w, f (20) -‐ l.1.N Now,

f(20) Finally, f (20) = 3(20) + 2 = 62. Finally, g(10) = 62 62‐- 11 = 61.

_ 22

iol 16 + (- 4) 32 7. ‐ 44))=: 16‐2%4)i : 3‐1 2 7. .~ f (J (2(- 4) = - 8 =

8.

Lookingat only when when Looking at the chart, chart, fJ (x) z = 44 only x= =3.3. So Sokk = 3.

-4 4..

15.

[I] We plug plug in values values to solve solve for a and and b.

lg] g(x) constant of 7 to [Q]Since g( x) just adds adds a constant every the maximum every value value of f(x), J (x), the maximum of g(x) must occur occur at the same must x-value as same x-value as the maximum maximum of f(x), J (x), namely namely x = 3. So, the maximum reached at the point point maximum of g(x) is reached at the (3,g(3)), since g(x) : Jf(x) 7, this this (3,g (3)), and and since g (x ) = (x) + 7, point is is (3,f(3) (3,/ (3) + point + 7).

Plugging in (0, (O, -‐ 22), ) , -‐22 = : a(0)2 + b = : b. So, Plugging b = -‐ 22.. Plugging Plugging in (1,3), (1, 3),

a(1)2+ 3 = “(1)2 +b 3= z a -‐ 2 = a 5=

16. .0 f ( J1(18) 8 ) =: \/18 = v'16 \/1‐ = z 4. 4. ✓18 ‐- 2 =

= /f(11) (11) =

So a = 55 and Soa x) = : 5x 5x22 -‐ 2. Finally, Finally, and Jf ((x)

\/11‐ /If=2

= \/§ 3. = J9 == 3.

= 4 -‐ 3 = Testingeach ff(18) (18) ‐- ff(11) (11) = = 1. Testing each

[J((3) 3 ) == 5(3)2 ‐ 22 == 43. 5(3)2 -

9. 0

2f(k) =8 2f (k) = 8 flJ (k) k) = Z 44

J

answer also answer choice, choice, f (3) is the only one one that that also

equals 1. equals l.

Plug in the and check. check. Plug the answer answer choices choices and

1 1 2 1 . . 1 f(§) _ (5) ‐ z,whlchlslessthan 2' J (1)= (1) = ~,which is less than 1 2

The answer (A). answer is (A).

311

CHAPTER 30 ANSWERS TO IBE EXERCISES

17. ~ factor g, I If If we we factor 3, we we get get 2 + 4x + 4 = (x + 2) 2 . Since g(x) is g(x) = x g(x )= x + 4 x + 4 ‐_ (x+2)2. Sinceg(x) is f (x)) shifted shifted k units units to the the left,

25. [[I each of the the answer answer choices When E Test each choices.. When x == -‐ 33,, Jf (x) = -‐22 according according to its graph , its graph, g(x) = ((-‐ 3 ++3)(‐‐3 3)( - 3 -‐ 1) 0.Inthis and g(x) 1) = 0. In this case , greater than than g(x). g(x ). When case, f ((x x ) is not n o t greater When and x == -‐ 22,, Jf ((x) x) ~ z 1.5 and g(x) =: ((-‐ 2 + 3 3)( this case, g(x) ) (-‐ 2 -‐ 1) 1) = z -‐ 33.. Ilnn this case, fJ((x) x ) >> g(x) so so we have have oour u r answer answer..

g(x) +kk)) g ( x )== J(x f(x+

(x+2)2=(x+k‐3)2 (x + 2)2 = (x + k - 3)2 + 22 =: xx + x+ + k -‐ 3 2= = k -‐ 3 5=k

26.

the value value of x that that makes makes the the substituted the substituted equal to O. 0. This value tells tells you you expression equal expression This value what the horizontal horizontal shift For choice what the shift is. For choice A,

Alternatively, this Alternatively, we we could've could’ve solved solved this and g(x) que stion by comparing question comparing J(x) flat) and g(x) to to 2 . The y= offfJ((x) - 3) 3 : xx2. The graph graph o x ) == (x ( x~ 3)22 is i s3 2, and units units to the the right right of y = : xx2, and the the graph graph of 2. g( x) = (x + the left g(x) + 2) 2)22 is is 2 unHs units to to the left of y = xx2. Therefore, the left left of J( x ). Therefore, g(x) g(x) is 5 unit unitss to the f (x)

2

= 5 makes makes 3x -‐ 2 equal equal to 0, so so the x= the 3

horizontal units to the right. For For horizontal shift shift is ~g units the right. 2 2 that value the shift shift is g choice B, that choice value is -‐ g,, so so the

3

the expres sion 2x -‐ 33 equal the shift the expression equal to 0, so so the shift is 3 ~5 units units to the the right. This is the answer . For right. This the answer. For

[Qj @ g( g(aa) = 6

choice D, the the shift shift is ~3 units units to the the left. left. choice

v13a =6 36 3a = 36

27. [Q] The key are "linear ": f is [E The key words words are ”linear function function": is a what straight line can a straight straight line. line. So So for what straight line can both f(4) 2 /f(5) be true? both f(2) / (2) 5 '.Sf[(3) (3) and and J( 4) 2: (5) be true? Only Take aa minute Only a a horizontal horizontal straight straight line. line . Take minute to think since f is a flat think that that through. through . Now, Now , since a flat line 10, then all values values of are 10, 10, line and and f/ (6) (6) = 10, then all offf are no matter value of x is. Therefore, no matter what what the the value Therefore, f mO)) = : 110. a f(

12 a = 12 E] Using the tabl table, g(‐1) : 2. Then, Then, 20. [Q] Using the e, g( - 1) =

/( 2) == 66. f (2) 21 21.

3

units left. For For choice units to the the left. choice C, x = ~3 makes makes

18. [Q] 2) cannot E) (1, (1,2) cannot be be on on the the graph graph of y since since an an x-value of 1 would result in division by 0. x-value would result division 0. 19.

[g For For horizontal horizontal shifts, the trick trick is to find shifts, the is to find

[]Ju g(c) = 5, then then c =

since 1 is the the only only 1 since input that gives an an output 5. Then, Then, input that gives output of 5. f(c) f(C) == fJ( (l1 )) == 33.‑

28.

Remember that 8JRemember that you you can can use use your your

calculator for graphing graphing.. The The graph graph of calculator for x ) = xx33 is ”centered" (0,0). Jf ((x) "centered" at at (0,0 ). The The graph graph of g(x) is ”centered” at (3, (3, -‐2). Comparing g(x) is "centered" at 2). Comparing these see that that g(x) g(x) these points points of reference, reference, we we can can see is shifted shifted 3 units to the and 22 units units units to the right right and downward downward from from [(x). J(x) . Therefore, Therefore, g(x) : J(x f ( x -‐ 3) -‐ 2, 2,which = -‐33 g(x) = which means means a =

From the the second second equation, = 20. So, So, 22.. []] From equation , f (a) =

f(a) a+ + 55 J( a) = -‐ 33a 20 == -‐3a 20 3a + 5 3a == -‐15 3a 15 a =z -‐ 5

and um a + +b The ssum bisis then then and b = -‐ 22.. The = -‐ 55.. -‐ 33 + + ((-‐ 22)) =

.f-)g(

: 3=) /( 2(3) 2 ( : 3- )1) ‐ 1=) /( f(5)=2.Weget 23. 8]g( (3) 5) = 2. We get from the the table. table. f (5) =‐2 2 from

f@)f(8) = 4(8) -‐ 33= Testing each each 24. [Q] J (8) = = 29. Testing answer choice choice to see see which which one one yields yields 29, we answer that g(8) 3(8) = 3(8) + 5 = 29. see that see

312

PANDA COLLEGE PANDA THE COLLEGE THE

9. they-intercept When x = 29. @ When = 0, y = = 9, 9, so so the y-intercept is 9. 3. is x-intercept the x-intercept so the 0, x = 3, so When y = O,x When

31.

slope of with aa slope line with a line function g(x) is a The function [g The draw g(x) If you a y-intercept and a l1 and y-intercept of k. If you draw g(x) with with the from k for possibilities different the the different possibilities from the an there's an that there’s see that you'll see choices, you’ll answer answer choices, when only when with J(x) points with intersection f (x) only intersection of 3 points below . shown below. as shown k = 1 as

y A

y

and 3 and base of 3 a base with a triangle with right triangle a right AAOB 6 AOB is a theorem a pythagorean theorem,, the pythagorean Using the height of 9. Using a height

A W2 0Bfi2z=AAB +o A0d2 + ¢ A §2 + 3§2 == AB 92 + s =zABfi2 90 3 ¢ E =: AAB B 3v'10 30.

32.

( - a, a), Plugging in (‐a,a), [I]Plugging

a= (‐a + l12 2 a)) + = aa(-

where J units up f from where up from graph of g is 4 units The graph [g The

a: - a 2 + 12 a = ‐a2+12 a2+a‐12=0 a2 + a - 12 = 0

and the x and slope off the slope is, because the of f is -‐ 22,, the but because is, but y intercepts will not increase by by the the same same not increase intercepts of g will So a ratio amount. increase in a ratio of 2:1. 50 They'll increase amount. They’ll the 4, the up by 4, shifted up when the gets shifted -in tercept gets theyy-intercept when x-intercept gets the right right by by 2. The The shifted to the gets shifted x-intercept n e w x-intercept x-intencept is therefore therefore 1I + 2 = 3. new solve for Another actually solve this is to actually do this way to do Another way form, -intercept form, the slope-intercept Using slope x-in tercept. Using the x-intercept. we get f(x) = ‐ 2 x + 2.Adding 4 to get the the get to 4 Adding 2. 2x J(x) get we equation of g, g(x) ‐ 2 x + 6. Setting Setting 6. + 2x = g(x) equation the g(x) = 00 and and solving solving for x to get get the g(x) x-intercept, get x = z 3. we get x-intercept, we

((a+ a + 44)(a ) ( a-~ 3 3)) == 0 a= ,3 = ‐-44,3 S i n c ea a>>0, 0 ,aa= = 33.. Since

313

ANSWERS TO THE EXERCISES CHAPTER 30 ANSWERS CHAPTER EXERCISES

Quadratics 17:Quadratics Chapter17: Chapter CHAPTER EXERCISE: CHAPTER

1.

x-intercepts. the x-intercepts. [g We factor factor to find the

6.

vertex. Since m is occurs at its vertex. always occurs always a upwards in a opens upwards positive, parabola opens the parabola positive, the with dealing we're means which shape, ”"U" U ” shape, which means we're dealing with Since vertex . Since its vertex. at its minimum at the parabola 's minimum the parabola's expanding vertex form form of a vertex us a gives us expanding f gives f(x) 1]= ( x-‐ m)2 ‐- mm,, m(x =m m [(x - m)2 - 1] J (x) = m[(x‐m)2‐ the Therefore, the vertex is at (m, ‐- m). Therefore, the vertex the m). occurs at (m, ‐- m parabola’s ). minimum occurs parabola 's minimum

3x - 10 = (x - 5) (x + 2 ) y= y=x2‐3x‐10=(x‐5)(x+2) x2 -

distance The distance and ‐- 22.. The are 5 and x-intercepts are The x-intercepts The them is 5 ‐- ((-‐ 22)) = between them between = 7. formula, quadratic formula, 2. [}] Using Using the the quadratic

X

=

✓ (4 ) 2 - 4(1)( 2) -‐ 44 fl±: ./(4)?‐4(1)(2) 2 (1) 2(1)

_ = _

v'8 -‐ 44±i \/§

the into the equation into the first equation 7. Substituting the 7. [}] Substituting second, second,

22

_‐4:t2\/§ _#- 4 ± 2v12

‐3 ex +cx x2 + - 3==x2

2

:=-‐ 22i ± \ / ..fi §.

2 0 + at + 33 x +ex+ 0 = x2

will have equations will The have ttwo wo system of equations The system two has above equation above has t w o the equation solutions if the solutions if have ttwo above to have equation above the equation For the solutions. For solutions. wo be must be 4ae, must b2 - 4ac, discriminant, b2‐ the discriminant, solutions, the solutions, positive. positive.

3 .‑ 3. []] 2a2‐7a+3:0 2a2 - 7a + 3 = 0 ((2a 2 a- ‐l1)(a )(a 0 - ‐3)3 )=: 0

c2‐ 0 e2 - 4(1)(3) > o

remember this , remember factoring this, trouble factoring had trouble you had If you quadratic the quadratic use the always use can always you can that you that

c2 0 12 > o e2 -‐ 12

1

.

0.5. = g,, or 0.5. 1, a = < 1,a Smee a < formula. formula. Since

2 > 12 > 12 cC2

2

4. 4.

Expanding everything, [iJExpandingeverything,

choices, only answer choices, the answer each of the Testing only Testing each than 12 bigger than value bigger gives a value (A), ‐- 44,, gives answer (A), answer when squared. when squared.

(2x‐3)2 +5 (2x - 3)2 = 44xx + 4x2‐12x+9=4x+5 4x2 - 12x + 9 = 4x + 5 4x2-16x+4:o 4x 2 - 16x + 4 = 0

treat the points , treat intersection points, the intersection 8. @ To find the equations. system of equations. as a system equations asa ttwo w o equations second , the second, into the equation into the first equation Substituting the Substituting

- 16 b . . b - - 4- = = ‐T is - z solutions IS_E the solutions sum of the The : 4. The sum

4:(x+2)2‐5 4 = (x + 2) 2 - 5 = (x + 2) 2 99=(x+2)2

a

5. 5.

parabola a parabola maximum of a minimum or maximum The minimum []JThe

get side to get left side the left the 8 to the Move the lg!Move

i±S3 = x + 2 xX :=‐ -5 5,, 1

the either use can either we can Now, l0x -‐ 8 = 0. N + 10x 3x + o w, we use the this case, formula or factor quadratic formula quadratic factor.. In this case, we'll we’ll =0 factoring: (x + with go w i t h factoring: + 4)(3x ‐- 2) =

points intersection points the intersection y-coordinates of the The y‐coordinates The the ttwo so the equation), so the first equation), (from the be 4 (from must be must wo (1, 4). and (1,4). (- 5, 4) and are (‐5,4) intersection are points of intersection points

b, b must be ~ . Since a > b,bmustbe orr xx == §.Smcea S o ,xx== -‐ 44 o So, = 16. ( - 4) 2 = = (‐4)2 b2 = and 172 ‐4 - 4 and

314

PANDA COLLEGE PANDA THE COLLEGE THE

9.

the (3, -‐ 88), vertex is at (3, the vertex Because the [g Because ) , the

14.

the Because the (C). Because (A) or (C). either (A) must be either answer must answer use can use we can (1,0 ), we through (1,0), passes through parabola passes parabola potential our test oout to test point to that point that ut o u r ttwo w o potential (C), we = x in plug we When answers. answers. When plug = l1 into into (C), (C). answer is (C). the answer that the 0, confirming that get y = 0,confirming get

0

the into the equation into Substitute the first equation Substitute the second, second, - 3 = ax2 + 4x - 4 ‐3=ax2+4x‐4 0=ax2+4x‐1 0 = ax2 + 4x - 1

the solution, the real solution, one real have one For the system to have the system real one only have should above equation equation above should have only one real discriminant, the discriminant, words, the other words, solution . In other solution. equal 0. must equal b2 - 4ac, must b2‐

t(S -‐ t), t2 = v = St equation 0 the equation 10. ) 2.5 ) From From the 5t -‐ t2 : t(5 Oand are 0 the t-intercepts that the can see we can we see that t-intercepts are and 5. 5. occurs at the maximum occurs the maximum Because the Because the vertex, vertex, average of the the average t-coordinate is the whose t-coordinate whose the two two maximum the in results 2.5 = t t-intercepts, t-intercepts, t = results the maximum graphing by graphing this by confirm this can confirm v. You can value of 1). value calculator . your calculator. on your the equation on the equation

2 0 4 (a)( - 1) = (4) (4)2 -‐4(a)(‐1) =0 1 6++ 4a 4 a== 00 16

4 6 4aa== ‐- 116

I

number of minimum number the minimum 11. ) 400 To find the that it so that must company m the company mattresses the mattresses u s t sell sell so money, set lose money, doesn't lose doesn’t set P = = 0.

a= = ‐- 4

15.

m2 120,000 = = 0o m2 -‐ 100m -‐ 120,000

Since value of J minimum value the minimum results f (x). Since results in the opens that parabola a offf is aparabola that opens the graph of the graph offf minimum of the minimum shape, the a ”"U" upwards U ” shape, upwards in a located at which is located the vertex, occurs vertex, which occurs at the xx =_-_baE=_- _- (24it) = 12. 12. Therefore, Therefore, the the 2 1 2 2a 2(l) units each produce 12 should produce manufacturer 12 units each manufacturer should unit. per unit. cost per the cost minimize the week week to minimize

(m -‐400)(m + 300) = 00 400)( m +300) m= = ‐300,400 - 300,400 number of the number sense for the make sense doesn't make Since it doesn’t Since negative , m = sold to be negative, mattresses : 400. If If mattresses sold above equation above the equation factoring the trouble factoring had trouble you had you and calculator and your calculator on your tough), graphing (it's tough), (it’s graphing on good both good are both quadratic formula the formula are the quadratic alternatives.. alternatives 12.

that value of x that the value looking for the We're looking [ill We're

02]

are x-intercepts, x-intercepts , bare and x = b 0 and Since xx = O 16. El Since and Oat f (x), the at x = 0 and speed, is 0 transfer speed, data transfer the data transfer x= : b. First, would the data transfer the data why would First, why transfer is the file transfer O? Well, the at x = Oat be 0 speed : 0? speed be just starting so no megabytes have been just starting so no megabytes have been

the number 10,000 The number 10,000 is they-intercept, the y-intercept, the [g The number the number total when x, the expenses when monthly expenses total monthly expenses of tables, is 0. We can assume these expenses these assume can tables, etc. salaries, etc. worker salaries, equipment, worker rent, equipment, be rent, to be

the speed transferred yet. Now why would would the speed Now why transferred yet. the file that the n S W e r iis s that best aanswer The best = bb?? The transfer no are no there are so there completed, so just completed, has just transfer has transfer - just to transfer‐just more left to data left megabytes of data more megabytes at stops at when it stops Owhen like be 0 would be speed would car's speed like a car's likely most likely Therefore, b most trip. Therefore, a trip. end of a the end the represents which the transfer the file transfer time at which the time represents the completed. completed.

b a t xx = bee 0Oat

square. First First the square. complete the need to complete We need 13. [[] We divide by -‐ 1, everything by divide everything

y = x2 - 6x - 20 -‐y=x2‐6x‐20 term by 2 to the middle divide the Now N o w divide middle term to get get -‐33 the -‐33 put and square that result to get 9. We put the get result that and square subtract the and subtract with x and parentheses with the parentheses inside the inside the 9 at end. the end. at the

17.

and 0 and g(x) and x = =O parabola and x) is aa parabola Since g( [g Since

x= = ~g is the the x-intercepts, x = its x-intercepts, are its = c are

= (x‐3)2‐20‐9 (X - 3)2- 20 - 9 -‐ y =

parabola’s along line along the line symmetry, the axis of symmetry, parabola's axis vertex the vertex case, the this case, vertex lies. In this the vertex which the which the since the occurs, since is where maximum occurs, the maximum where the parabola an downwards in an opens downwards parabola opens

back everything back N o w simplify multiply everything and multiply simplify and Now by 1. by -‐1.

y: ‐(x‐3)2+29

upside-down U ” shape. %is the is the Therefore, ~ shape . Therefore, upside-down “"U" time which the a s at a was speed w transfer speed data transfer the data time at which maximum maximum..

315

CHAPTER 30 ANSWERS TO THE EXERCISES

18.

[Q]One of the x-intercepts

is 3. Since the the at the must lie at 5, must vertex, 5, the vertex, x-coordinate of the x-coordinate other the other x-intercepts, the the ttwo midpoint of the midpoint w o x-intercepts, us giving us 7, giving Therefore , k = 7, x-intercept is 7. Therefore, x-intercept the plug in the now can n We can 3)( x -‐ 7). We = a(x -‐ 3)(x y= o w plug solve for a. point to solve as a point vertex as vertex ‐- 332 2 = a(5 -‐ 3)( 3)(55 -‐ 7)

a(2)( - 2) ‐- 332 2 = a(2)(‐2) ‐- 332 2 := ‐- 44a a a a == 8B

19.

both into both (3, k) into point (3, the point Substituting the [I] Substituting equations, equations,

k :=22(3) ( 3 ) +b

k=(3)2+3b+5 k = (3)2 + 3b + 5 Substituting the equations. Substituting system of equations. This is a system This the second, the second, into the equation into first equation

2 ( 3+ ) +bb=: (3)2+3b+5 (3)2 + 3b + 5 2(3) 6+ + b := 9 + + 3b 3b + 5

+bb 2=33b 6+ b ++114 4 : 22b b -‐ ‐88= : ‐- 4 b= equation, first equation, the first From the From

k = 6+ + bb == 66 -‐ 4 = 2

316

THE THE COLLEGE COLLEGE PANDA PANDA

Chapter 18: Synthetic Chapter Synthetic Division Division CHAPTER CHAPTER EXERCISE: EXERCISE:

[g

1. l.

5. [TIzz ‐- 11is only ifif the 5. is aa factor factor only the polynomial polynomial yields O yields 0 when when z2 = = 1 (the (the remainder remainder theorem). theorem). Therefore, can set set up up an an equation. equation. Therefore, we we can

4

2 I4x x ‐ 2 4x 4x 4 x -‐ 8 8

2(1)3‐kx(1)2+5x(1)+2x‐2=0 2(1) 3 - kx (1) 2 + 5x (l ) + 2x - 2 = 0 2 ‐kx+5x+2x‐2=O 2 - kx + 5x + 2x - 2 = 0

X -

l

8 3

+ 7x 7 x:= 0 -‐ kkxx + 0 From can see see that that k = From here, here, we we can 2 7. 7'

This This result result can can be be expressed expressed as as4 + _‐xfB_2 4+ x- 2

z-

‘

7'‑

7. []]

3x 3 x ++ 1

2 x++11I6x 6 2x 2+ +5x 5 x++ 22 2x 6x 2

6x

6. [2J By the the remainder 6. E By remainder theorem, the remainder theorem , the remainder 2 is(‐4)2+2(‐4)+1=16‐8+l=9. is (- 4) + 2(- 4) + 1 = 16 - 8 + 1 = 9.

‘

xX

+ + 33xx

‐

-

3X2 _ 21 3x 2 - 2x

2x + 11

‐ 6x ‐ 4 6x

11 This This result result can c a n be be expressed as expressed as . 1 3x + 1 1 + -m, Wthh Q Q= = 3x 3x + + 1. 1. , from from which 2x + l

-‐

,

expression by 2x -‐ 1l and expression and write write the the result result in in

the form of the form of Dividend : Quotient Divisor + Remainder. Dividend = Quotient xx Divisor + Remainder. 2x + + 11 2x 2 -l-5 2 X- ‐ 11j 4x 4X 2x +

!iD

‘ ,

2x

.

'

_

_

. _

8. to divide diVide the the 8. El This This question question 15 is asking asking you you to expression by xx + + 11 and and write the result in the expression by write the result in the form form of Dividend : Quotient + Remainder Remainder.. Dividend = Quotient x>< Dividend Remainder, x Divisor Divisor + Remainder, where ax + b where is the is the the bis the quotient quotient and and c c is remainder remainder..

14.

be be a factor factor of p(x) ifif p

+ 6 + 2 xX -‐ 221xx 2 + 4x 4 x- - 99 x X

+

xx22 -‐ 22x x 6x x - ‐ 9 6 6xx -‐ 112 6 2 3

Therefore, x2 + 4x - 9 = (x Therefore,x2+4x‐9= ( x ++ 66)() (xx-‐ 22)) + + 33..

Finally, a= F i n a l l y, a =1,l ,bb== 6, 6 ,cc= =3 a n daa++bb++ cc ==110. 0. 3,,and

10. ~ Using Using the the remainder remainder theorem, theorem, p(2) = =0 means that that x -‐ 2 is a a factor factor of p(x). means 11. ~ Use the remainder remainder theorem theorem to test Use the test each each option for a a remainder remainder of 0. option 23 + 22 - 5(2) + 3 = p(2) = =23+22‐5(2)+3 = 55.. 13 + 12 - 5(1) + 3 = 0. p(1) = : 13+12‐5(1)+3=0.

p(- 3) = (‐3)3 (- 3)3 + (- 3)2 -‐ 5(-3) 5(- 3) + p(‐3) + (‐3)2 + 3 =z o. 0. Therefore, divisible by x -‐ 1 and Therefore, p(x) is divisible and x + 3.

12.

[ill ItIf p(x) is divisible divisible by x -‐ [E]

2, then then p(2) must equal equal 0 (the (the remainder remainder theorem). must theorem). Testing each each answer answer choice, choice, only Testing only choice choice (D) results in 0 when results : 2. when x =

13. ~ Using Using the the remainder remainder theorem, theorem, we we can can set set up a a system up system of equations equations.. When When the the polynomial is divided divided by by x ‐- 1or polynomial 1,the 1 or x + 1, the remainder is 0, which remainder which means means that that if we we let let denote the the polynomial, polynomial, p(1) = 00 and p(x) denote and p(- 1)== 00. p(‐1) a(1)4 + b(1) 3 - 3(1)2 + 5(1) a(1)4+b(1)3‐3(1)2+5(1) =o =0 { a(‐1)4 a(- 1)4 + b(- 1)3 -‐ 3(‐1)2 + 5(5(‐1) =o + b(‐1)3 3(- 1)2 + 1) = 0 a+ + b ‐- 3 + + 5

[I] From From the the remainder remainder theorem, theorem, 3x -‐

= 0 =

{ a ‐- b ‐- 3 ‐- 5 z= 0 Adding the the equations equations together, Adding together,

2 a- ‐ 66=: 0 2a a= 3

318

(g) (1)= 0.0.

11 must must

THE PANDA COLLEGE PANDA THE COLLEGE

Chapter19:Complex Numbers CHAPTER EXERCISE:

.I(5‐3i)‐(‐2+5i)=5‐3i+2‐5i=7‐8i = 5 - 3i + 2 - Si = 7 - Bi i (ii(i++11)) == ii22++i =i :-‐1l + ii 2. I[[j 1. [g (5 - 3i) - (- 2 + Si)

.Ii4+3i2+2=1‐3+2:0 + 3i + 2 = 1 - 3 + 2 = 0 5 ( ‐i)i )++ 6(1) 6 ( 1 =) :44-‐ 22ii 0 2 + 3i + 4i + 5i + 6i = 22 ++33ii ++44(-( ‐ 11))++ 5(4...2+3i+4i2+5i3+6i4 2

3. [g i4

2

3

4

S n daa++bb= : 22.. Sooaa=: 44,, bb == ‐-22,, aand

0

(6 + 2i)( 2 + Si) = 12 + 30i + 4i + 10i2 = 12 + 34i + 10(- 1) = 2 + 34i 5.. (6+21)(2+5i)=12+30i+4i+10i2=12+34i+10(‐1)=2+34i =22.. Therefore,a Therefore, a 2

lli i Bii = : 33ii ++66 ‐- 110 0+8 = -‐ 44 ++11 [g 3(i + 2) - 2(5 - 4i) = 6..@3(i+2)‐2(5‐4i)

[[l 3i(i + 2) - i(i - 1) = 3i2 + 6i - i2 + i = -‐ 3 ++ 66ii-‐ ((-‐ 11))++ ii = ‐- 22++77ii 7..I3i(i+2)‐i(i‐1)=3i2+6i‐i2+i= (1)23. i = i @]i93 = (i4)23. i = (1)23-i=i 8..Ei93=(i4)23-i:

0 (3 - i) 2 = 32 - 6i + ;2 = 9 - 6i - 1 = 8 - 6i .-(3‐i)2=32‐6i+i2=9‐6i‐1=8‐6i

9.

10.

i2 - i4 = 0 Deal with the exponents first: (- i)2 - (- i)4 = 1'2‐i4 1Z'Dealwiththeexponentsfirst:(‐i)2‐(‐i)4 = ‐-

11.

(5 - 2i)( 4 - 3i) = 20 - 15i - Bi+ 6i2 = 20 - 23i - 6 = 14 - 23i [[j E(5‐2i)(4‐3i)zzo‐151‐81+6i2:20‐231‐6=14‐23i

1 -‐ 11=: -‐22

1 1 1 1 = ! - 1+1 = ! + ..!_ !1 + ..!_ .'Z|7+i‐2+i‐4_7‐1+1_7 G\l i i i4 i2 ~ i

12 12.

~ = ‐i- i = 1L= getygz I

byy i ttoo get~·~ Now bottom b and bottom top and both top multiply both Now multiply I

.(1-3i)(3‐1’)_3‐i‐9i‐1~3i2

I

_ ‐ -1lOi 3_ 3 -‐ 1lOi 3 - lOi + 3 0 i-‐ 3 0 i -2_ ‐- 1/. !Al (1 - 3i) (3 - i) _ 3 - i - 9i + 3;2 _ 3‐101'+3i2 - -- ---,----,-- ----=--- 2 - -----=· -13 L.:2J-2 10 9- i 9 - 3i + 3i - i (3 - i) . ((33+ + ii)) (3‐i)_9‐3i+3i‐i2: 9-12 “ 9 9‐ -( ‐(-11)) 10 _

14 .

‐

3i2 _

_

‐

_

2 2 .(2‐i).(2‐i)_22‐2i‐2i+i2_4‐4i+i2_4‐4i‐1_3~4i (2 - i) ' (2 - i) = 2 - 2i - 2i + i2 = 4 - 4i + i = 4 - 4i - 1 = 3 - 4i = 0 2 2

+ 22i i-‐ i 2 _ ((22++ ii)) ((22- ‐ i) i ) _ 44 ‐- 22ii +

4- i 4‐12

1) ) " ‘ 44‐- ( (‐- 1

5 5

!i ~ -_ 2 § “5 5 5 5

IThe

i)(1 + i). i). denominator is (1 -‐ i)(1 common denominator 15.. [[] The common 4ii++ i 22) )++(2 ((44++ ii)( ) (11++ ii)) (2 ( 2-‐ ii)( ) (11-‐ ii)) _ (4 ( 4++ii ++ 4 ( 2- ‐ 2i 2 i-‐ ii++i 2i2 )) _ 44 + Si 5 i- ‐ 1l ++ 2 -‐ 33ii -‐ 1 _ 44 ++ 22ii 2 2== 11++ ii -‐ ii ‐- i i2 ((1l ‐- i )i)(1 ( 1 + ii)+( ) (ll ‐- i i)(1 ) ( 1++i)i )=‐ ‘ 1 -(‐ ( ‐ 11)) ' 2

: 22 ++ ii =

319

CHAPTER 30 ANSWERS ANSWERS TO TO THE THE EXERCISES EXERCISES CHAPTER 30

Chapter Chapter20: Absolute Value Value CHAPTER CHAPTEREXERCISE: EXERCISE:

1.‑

1. []]

1|f(1)| / (l ) I = 1- 2(1) 2 - 3 (1) + 11 = l- 4 1 = 4 =l‐2(1)2‐3(1)+1|:|_4|=4

2. The best to solve question is is trial and error error.. If = 1, 1, which which is is not not 2. ~ The best way way to solve this this question trial and If x x := 3, 3, for for example, example, 1|2 2 -‐ 33|J = greater than result indicates that we we should should try try larger we continue continue to o u r way greater than 5. This This result indicates that larger numbers. numbers . If If we to work work our way up, arrive at at the minimum possible possible value value xx = = 8, which results results in in J2 |2 -‐ BJ 8| = : 6. 6. up, we we would would arrive the minimum 8, which 3.

can equal equal -‐55 (when []J Only Only the the expression expression in answer answer (B) can (when x

= 11 or Because the absolute value value of = or 3). Because the absolute of anything is always greater than or equal to O, the other answer choices can never reach ‐ 5 . anything is always greater than or equal to 0, the other answer choices can never reach - 5.

Recall that that the graph of y = Jx [x]l is aa V-shape V‐shape centered centered at The graph graph pictured pictured is 4. []] Recall the graph at the the origin. origin. The is also also V‐shaped but but converges converges at at y = = -‐ 22,, which which means means it has has shifted w o units V-shaped shifted ttwo units down. down. Therefore, Therefore, the the equation equation of the the graph : lx le l -‐ 2. Note Note that |x -‐ 21 2| shifts shifts the w o units units to the O T ttwo w o units units graph is y = that y = Ix the graph graph ttwo the right, right , N NOT down down.. 5. 5. [ill IE Test each choices, making making sure sure to include negative possibilities. possibilities. For For example, the each of the the answer answer choices, include the the negative example , the answer is not answer n o t (A) because z 2 or -‐ 2, 2, Ix [x -‐ 33]J is n o t greater than 10. However, However, Ix [x -‐ 33]1is greater greater because when when x = not greater than than 10 when x = than 10 when : -~8. 8.

6.

@]Smart Smart trial trial and and error error is the the fastest fastest way way to find find the the bounds The lower lower bound for xxisis -‐88 and and the bounds for x. The bound for the upper upper bound bound is -‐ 44.. There There are are 5 integers integers between between -‐44 and do this and -‐88 (inclusive). (inclusive). If we wanted wanted to do this problem problem more more mathematically, mathematically, we we could could set set up up the the following following equation: equation: < -‐3 3