Chapter 3 - WHS SALISBURY - Homewhssalisbury.weebly.com/.../8/112801805/hscc_alg2_wsk_03.pdf−8 =...

Copyright © Big Ideas Learning, LLC Algebra 2 89All rights reserved. Worked-Out Solutions

Chapter 3

Chapter 3 Maintai ning Mathematical Profi ciency (p. 91)

1. √—

27 = √—

9 ⋅ 3

= √—

9 ⋅ √—

3

= 3 √—

3

2. − √—

112 = − √—

16 ⋅ 7

= − √—

16 ⋅ √—

7

= −4 √—

7

3. √—

11

— 64

= √

— 11 —

√—

64

= √

— 11 —

8

4. √—

147

— 100

= √—

147 —

√—

100

= √—

49 ⋅ 3 —

10

= √—

49 ⋅ √—

3 —

10

= 7 √—

3 —

10

5. √—

18

— 49

= √

— 18 —

√—

49

= √—

9 ⋅ 2 —

7

= √—

9 ⋅ √—

2 —

7

= 3 √

— 2 —

7

6. − √—

65

— 121

= − √

— 65 —

√—

121

= − √—

65 —

11

7. − √—

80 = − √—

16 ⋅ 5

= − √—

16 ⋅ √—

5

= −4 √—

5

8. √—

32 = √—

16 ⋅ 2

= √—

16 ⋅ √—

2

= 4 √—

2

9. x2 − 36 = x2 − 62

= (x + 6)(x − 6)

So, x2 − 36 = (x + 6)(x − 6).

10. x2 − 9 = x2 − 32

= (x + 3)(x − 3)

So, x2 − 9 = (x + 3)(x − 3).

11. 4x2 − 25 = (2x)2 − 52

= (2x + 5)(2x − 5)

So, 4x2 − 25 = (2x + 5)(2x − 5).

12. x2 − 22x + 121 = x2 − 2(x)(11) + 112

= (x − 11)2

So, x2 − 22x + 121 = (x − 11)2.

13. x2 + 28x + 196 = x2 + 2(x)(14) + 142

= (x + 14)2

So, x2 + 28x + 196 = (x + 14)2.

14. 49x2 + 210x + 225 = (7x)2 + 2(7x)(15) + 152

= (7x + 15)2

So, 49x2 + 210x + 225 = (7x + 15)2.

15. Rewrite ax2 + 8x + c as ( √—

a )2x2 + 8x + ( √— c )2. For the

trinomial to be a perfect square, 2 √—

ac = 8. Therefore,

a = 16 and c = 1 or a = 4 and c = 4 or a = 1 and c = 16.

Chapter 3 Mathematical Practices (p. 92)

1. The graphing window has a limited number of pixels, and if

the height and width are not in the correct ratio, the graphs

will not connect.

2.

15

−10

10

−15

y = √x2 − 1.5

−15 < x < 15 and −10 < y < 10; The graph has a width-

height ratio of 3:2 and shows the graph touching the x-axis.

3.

15

−10

10

−15

y = √x − 2.5

−15 < x < 15 and −10 < y < 10; The graph has a width-


90 Algebra 2 Copyright © Big Ideas Learning, LLC Worked-Out Solutions All rights reserved.

Chapter 3

4.

7

−5

5

−7

y = ±√−x2 + 12.25

−6.85 < x < 6.85 and −4.56 < y < 4.56; The graph has a

width-height ratio of approximately 3:2 and shows the graph

touching the x-axis.

5.

7

−5

5

−7

y = ±√−x2 + 20.25

−8.45 < x < 8.45 and −5.63 < y < 5.63; The graph has a

width-height ratio of approximately 3:2 and shows the graph

touching the x-axis.

6.

7

−5

5

−7

y = ±√− x2 + 3.062514√

−6.3 < x < 6.3 and −4.2 < y < 4.2; The graph has a width-


7.

7

−5

5

−7

y = ±√−4x2 + 20.25

−9.6 < x < 9.6 and −6.4 < y < 6.4; The graph has a width-


3.1 Explorations (p. 93)

1. a. f (x) = x2 − 2x = x(x − 2); The graph is E because the

parabola opens up and has x-intercepts 0 and 2.

b. f (x) = x2 − 2x + 1 = (x − 1)(x − 1); The graph is D

because the parabola opens up and has x-intercept 1.

c. The graph is C because the parabola opens up and the

y-intercept is 2.

d. f (x) = −x2 + 2x = −x (x − 2); The graph is A because

the parabola opens down and has x-intercepts 0 and 2.

e. f (x) = −x2 + 2x − 1 = −(x2 − 2x + 1) = −(x − 1) (x − 1); The graph is F because the parabola

opens down and has x-intercept 1.

f. The graph is B because the parabola opens down and the

y-intercept is −2.

2. a. Based on Exploration 1, the real solutions are x = 0 and

x = 2.

b. Based on Exploration 1, the real solution is x = 1.

c. Based on Exploration 1, there is no real solution.

d. Based on Exploration 1, the real solutions are x = 0 and

x = 2.

e. Based on Exploration 1, the real solution is x = 1.

f. Based on Exploration 1, there is no real solution.

3. Graph the function. If there are two x-intercepts, then there

are two solutions. If there is one x-intercept, then there is one

solution. If there are no x-intercepts, then there is no

real solution.

4. By graphing the related function, the graph has two

x-intercepts. The solutions of the equation are x = −2

and x = −1.

3.1 Monitoring Progress (pp. 94–98)

1. The equation is in standard form. Graph the related function

y = x2 − 8x + 12.

x

y

4

2

−2

−4

4

The x-intercepts are 2 and 6. The solutions, or roots, are

x = 2 and x = 6.


y = 4x2 − 12x + 9.

x

y

4

6

2

42−2

The x-intercept is 1.5. The solution, or root, is x = 1.5.


Chapter 3

3. Rewrite the equation in standard form − 1 — 2 x2 + 6x − 20 = 0.

Graph the related function y = − 1 — 2 x2 + 6x − 20.

x

y

−8

−12

−4

8 124

The graph does not intersect the x-axis. So, the equation has

no real solution.

4. 2 — 3 x2 + 14 = 20

2 —

3 x2 = 6

2x2 = 18

x2 = 9 x = ± √

— 9

x = ±3

The solutions are x = 3 and x = −3.

5. −2x2 + 1 = −6

−2x2 = −7

x2 = 7 — 2

x = ± √—

7 —

2

x = ± √

— 7 —

√—

2

x = ± √

— 7 —

√—

2 ⋅

√—

2 —

√—

2

x = ± √

— 14 —

2

The solutions are x = √

— 14 —

2 and x = −

√—

14 —

2 .

6. 2(x − 4)2 = −5

(x − 4)2 = − 5 — 2

The square of a real number cannot be negative. So, the

equation has no real solution.

7. x2 + 12x + 35 = 0 (x + 5)(x + 7) = 0 x + 5 = 0 or x + 7 = 0 x = −5 or x = −7

The solutions are x = −5 and x = −7.

8. 3x2 − 5x = 2 3x2 − 5x − 2 = 0 (3x + 1)(x − 2) = 0 3x + 1 = 0 or x − 2 = 0

x = − 1 — 3 or x = 2

The solutions are x = − 1 — 3 and x = 2.

9. To fi nd the zeros of the function, fi nd the x-values for which

f (x) = 0.

x2 − 8x = 0 x(x − 8) = 0 x = 0 or x − 8 = 0 x = 0 or x = 8 The zeros of the function are x = 0 and x = 8.


f (x) = 0.

4x2 + 28x + 49 = 0 (2x + 7)2 = 0 2x + 7 = 0

x = − 7 — 2

The zero of the function is x = − 7 — 2 .

11. When the initial annual cost is $21, the revenue function is

R(x) = −2000(x − 24)(x + 21). The zeros of the function

are 24 and −21, with an average of 24 + (−21)

— 2 = 1.5.

To maximize revenue, each subscription should cost

$21 + $1.50 = $22.50. So, the maximum annual revenue

is R(1.5) = −2000(1.5 − 24) (1.5 + 21) = $1,012,500.

12. If the height the egg container is dropped from changes to

80 feet, the function in part (a) changes to

h(t) = −16t2 + 80. The amount of time it will take to hit the

ground will also change.

h = −16t2 + 80

0 = −16t2 + 80

−80 = −16t2

5 = t2

± √—

5 = t 2.23 ≈ t So, it will take about 2.23 seconds to hit the ground. Part (b)

will not change because the difference between the function

in the example and h(t) is a vertical translation.


Chapter 3

3.1 Exercises (pp. 99–102)

Vocabulary and Core Concept Check

1. First, make a graph of the function related to the

equation, and then determine the x-intercepts of the graph.

If there are two x-intercepts, then these are the two solutions;

if there is one x-intercept, then there is only one solution;

and if there are no x-intercepts, then there is no solution of

the equation.

2. The question “What is the y-intercept of the graph of

y = (x + 5) (x − 2)?” is different than the other three. First

the answer to the three questions that are the same can be

found by fi nding roots of 10 − x2 = 3x.

10 − x2 = 3x

0 = x2 + 3x − 10

0 = (x + 5)(x − 2)

x + 5 = 0 or x − 2 = 0 x = −5 or x = 2 So, the answer to the three that are the same is −5 and 2. For

the one that is different, the y-intercept is −10.

Monitoring Progress and Modeling with Mathematics


y = x2 + 3x + 2.

x

y

4

6

2

42−2−4

The x-intercepts are −1 and −2. The solutions, or roots, are

x = −1 and x = −2.


y = −x2 + 2x + 3.

x

y4

−4

−2

42−2−4

The x-intercepts are 3 and −1. The solutions, or roots, are

x = 3 and x = −1.


y = x2 − 9.

x

y

−4

2−2

The x-intercepts are −3 and 3. The solutions, or roots, are

x = −3 and x = 3.

6. Rewrite the equation in standard form.

−8 = −x2 − 4 x2 − 4 = 0 Graph the related function y = x2 − 4.

x

y1

−5

31−1−3


x = −2 and x = 2.


8x = −4 − 4x2

4x2 + 8x + 4 = 0 Graph the related function y = 4x2 + 8x + 4.

x

y4

−2

2−2

The x-intercept is −1. The solution, or root, is x = −1.


3x2 = 6x − 3 3x2 − 6x + 3 = 0 Graph the related function y = 3x2 − 6x + 3.

x

y

2

−2

42−2

The x-intercept is 1. The solution, or root, is x = 1.


Chapter 3


7 = −x2 − 4x

0 = −x2 − 4x − 7 Graph the related function y = −x2 − 4x − 7.

x

y

−6

−8

−2

4−4−8

There is no x-intercept. The equation has no real solution.


2x = x2 + 2 0 = x2 − 2x + 2 Graph the related function y = x2 − 2x + 2.

x

y

2

2−2



1 —

5 x2 + 6 = 2x

1 — 5 x2 − 2x + 6 = 0

Graph the related function y = 1 — 5 x2 − 2x + 6.

x

y

12

4

8 124−4



3x = 1 — 4 x2 + 5

0 = 1 — 4 x2 − 3x + 5

Graph the related function y = 1 — 4 x2 − 3x + 5.

x

y4

2

−4

−2

8 124

The x-intercepts are 2 and 10. The solutions, or roots,

are x = 2 and x = 10.

13. s2 = 144

s = ± √—

144

s = ±12

The solutions are s = 12 and s = −12.

14. a2 = 81

a = ± √—

81

a = ±9

The solutions are a = 9 and a = −9.

15. (z − 6)2 = 25

z − 6 = ± √—

25

z − 6 = ±5

z = 6 ± 5 The solutions are z = 1 and z = 11.

16. (p − 4)2 = 49

p − 4 = ± √—

49

p − 4 = ±7

p = 4 ± 7 The solutions are p = −3 and p = 11.

17. 4(x − 1)2 + 2 = 10

4(x − 1)2 = 8 (x − 1)2 = 2 x − 1 = ± √

— 2

x = 1 ± √—

2

The solutions are x = 1 − √—

2 and x = 1 + √—

2 .


Chapter 3

18. 2(x + 2)2 − 5 = 8 2(x + 2)2 = 13

(x + 2)2 = 13 —

2

x + 2 = ± √—

13

— 2

x + 2 = ± √

— 13 —

√—

2

x + 2 = ± √—

26 —

2

x = −2 ± √—

26 —

2

The solutions are x = −2 − √—

26 —

2 and x = −2 + √

— 26 —

2 .

19. 1 —

2 r2 − 10 = 3 —

2 r2

−r2 − 10 = 0 −r2 = 10

r2 = −10

The square of a real number cannot be negative. So, the

equation has no real solution.

20. 1 —

5 x2 + 2 = 3 —

5 r2

− 2 — 5 x2 + 2 = 0

− 2 — 5 x2 = −2

x2 = 5 x = ± √

— 5

The solutions are x = − √—

5 and x = √—

5 .

21. A: −x2 − 6x − 8 = 0 −(x2 + 6x + 8) = 0 −(x + 2)(x + 4) = 0 x + 2 = 0 or x + 4 = 0 x = −2 or x = −4

B: 0 = (x + 2)(x + 4)

x + 2 = 0 or x + 4 = 0 x = −2 x = −4

E: 4(x + 3)2 − 4 = 0 4(x + 3)2 = 4 (x + 3)2 = 1 x + 3 = ±1

x = −3 ± 1 x = −4 or −2

Equations A, B, and E all have solutions x = −2 and

x = −4, which is the same as the roots given by graph.

22. Solve the given equation.

( x − 3 — 2 )

2

= 25 —

4

x − 3 — 2 = ± √—

25

— 4

x − 3 — 2 = ± 5 —

2

x = 3 — 2 ± 5 —

2

x = 4 or x = −1

The solutions, or roots, are x = 4 and x = −1. Graph B is the

only graph that has x = 4 and x = −1 as the x-intercepts.

23. The ± was not used when taking the square root.

2(x + 1)2 + 3 = 21

2(x + 1)2 = 18

(x + 1)2 = 9 x + 1 = ±3

x = −1 ± 3 The solutions are x = 2 and x = −4.

24. The square root of −4 is not a real number.

−2x2 − 8 = 0 − 2x2 = 8 x2 = −4

The equation has no real solution.

25. a. Sample answer: The equation x2 = 4 has two solutions,

x = 2 and x = −2.

b. The equation x2 = 0 has one solution.

c. Sample answer: The equation x2 = −4 has no real

solutions.

26. 5x2 − 4 = x2 − 4 4x2 − 4 = −4

4x2 = 0 Equation B is the only equation that has one solution because

it is the only equation, when simplifi ed, of the form ax2 = 0.

27. 0 = x2 + 6x + 9 0 = (x + 3)2

x + 3 = 0 x = −3

So, the solution of the equation is x = −3.

28. 0 = z2 − 10z + 25

0 = (z − 5)2

z − 5 = 0 z = 5 So, the solution of the equation is z = 5.


Chapter 3

29. x2 − 8x = −12

x2 − 8x + 12 = 0 (x − 2)(x − 6) = 0 x − 2 = 0 or x − 6 = 0 x = 2 or x = 6 So, the solutions of the equation are x = 2 and x = 6.

30. x2 − 11x = −30

x2 − 11x + 30 = 0 (x − 5)(x − 6) = 0 x − 5 = 0 or x − 6 = 0 x = 5 or x = 6 So, the solutions of the equation are x = 5 and x = 6.

31. n2 − 6n = 0 n(n − 6) = 0 n = 0 or n − 6 = 0 n = 0 or n = 6 So, the solutions of the equation are n = 0 and n = 6.

32. a2 − 49 = 0 a2 = 49

a = ± √—

49

a = ±7

So, the solutions of the equation are a = −7 and a = 7.

33. 2w2 − 16w = 12w − 48

2w2 − 28w + 48 = 0 w2 − 14w + 24 = 0 (w − 2)(w − 12) = 0 w − 2 = 0 or w − 12 = 0 w = 2 or w = 12

So, the solutions of the equation are w = 2 and w = 12.

34. −y + 28 + y2 = 2y + 2y2

0 = y2 + 3y − 28

0 = (y + 7)(y − 4)

y + 7 = 0 or y − 4 = 0 y = −7 or y = 4 So, the solutions of the equation are y = −7 and y = 4.

35. The equation that represents the area of the rectangle is

36 = x(x + 5). Solve the equation.

36 = x(x + 5)

36 = x2 + 5x

0 = x2 + 5x − 36

0 = (x + 9)(x − 4)

x + 9 = 0 or x − 4 = 0 x = −9 or x = 4 The solutions of the equation are x = −9 and x = 4, but only

x = 4 is valid.

36. The equation that represents the area of the circle is

25π = π(x + 3)2. Solve the equation.

25π = π(x + 3)2

25 = (x + 3)2

± √—

25 = x + 3 ±5 = x + 3 −3 ± 5 = x The solutions of the equation are x = −8 and x = 2, but

only x = 2 is valid.

37. The equation that represents the area of the triangle is

42 = 1 — 2 (x + 3)(2x + 8). Solve the equation.

42 = 1 — 2 (x + 3)(2x + 8)

84 = (x + 3)(2x + 8)

84 = 2x2 + 14x + 24

0 = 2x2 + 14x − 60

0 = x2 + 7x − 30

0 = (x − 3)(x + 10)

x − 3 = 0 or x + 10 = 0 x = 3 or x = −10

The solutions of the equation are x = 3 and x = −10, but

only x = 3 is valid.

38. The equation that represents the area of the trapezoid is

32 = 1 — 2 x [ (x + 2) + (x + 6) ] . Solve the equation.

32 = 1 — 2 x [ (x + 2) + (x + 6) ]

64 = x [ (x + 2) + (x + 6) ]

64 = x(2x + 8)

64 = 2x2 + 8x

0 = 2x2 + 8x − 64

0 = x2 + 4x − 32

0 = (x + 8)(x − 4)

x + 8 = 0 or x − 4 = 0 x = −8 or x = 4 The solutions of the equation are x = −8 and x = 4, but only

x = 4 is valid.

39. The equation can be factored, so solve by factoring.

u2 = −9u

u2 + 9u = 0 u(u + 9) = 0 u = 0 or u + 9 = 0 u = 0 or u = −9

The solutions of the equation are u = 0 and u = −9.


Chapter 3

40. The equation can be written in the form u2 = d, so solve by

using square roots.

t2

— 20

+ 8 = 15

t2

— 20

= 7

t2 = 140

t = ± √—

140

t = ± 2 √—

35

The solutions of the equation are t = −2 √—

35 and t = 2 √—

35 .


using square roots.

−(x + 9)2 = 64

(x + 9)2 = −64

There is no real solution.


using square roots.

−2(x + 2)2 = 5

(x + 2)2 = − 5 — 2

There is no real solution.


using square roots.

7(x − 4)2 − 18 = 10

7(x − 4)2 = 28

(x − 4)2 = 4 x − 4 = ± √

— 4

x − 4 = ±2

x = 4 ± 2 The solutions of the equation are x = 2 and x = 6.


t2 + 8t + 16 = 0 (t + 4)2 = 0 t + 4 = 0 t = −4

The solution of the equation is t = −4.


x2 + 3x + 5 — 4 = 0

4x2 + 12x + 5 = 0 (2x + 1)(2x + 5) = 0 2x + 1 = 0 or 2x + 5 = 0

x = − 1 — 2 or x = − 5 —

2

So, the solutions of the equation are x = − 1 — 2 and x = − 5 —

2 .


using square roots.

x2 − 1.75 = 0.5

x2 = 2.25

x = ± √—

2.25

x = ±1.5

The solutions of the equation are x = −1.5 and x = 1.5.


g(x) = 0.

x2 + 6x + 8 = 0 (x + 2)(x + 4) = 0 x + 2 = 0 or x + 4 = 0 x = −2 or x = −4

The zeros of the function are x = −2 and x = −4.


f (x) = 0.

x2 − 8x + 16 = 0 (x − 4)2 = 0 x − 4 = 0 x = 4 The zero of the function is x = 4.

49. To fi nd the zeros of the function, fi nd the x-values for

which h(x) = 0.

x2 + 7x − 30 = 0 (x + 10)(x − 3) = 0 x + 10 = 0 or x − 3 = 0 x = −10 or x = 3 The zeros of the function are x = −10 and x = 3.


g(x) = 0.

x2 + 11x = 0 x(x + 11) = 0 x = 0 or x + 11 = 0 x = 0 or x = −11

The zeros of the function are x = 0 and x = −11.


f (x) = 0.

2x2 − 2x − 12 = 0 x2 − x − 6 = 0 (x − 3)(x + 2) = 0 x − 3 = 0 or x + 2 = 0 x = 3 or x = −2

The zeros of the function are x = 3 and x = −2.


Chapter 3


f (x) = 0.

4x2 − 12x + 9 = 0 (2x − 3)2 = 0 2x − 3 = 0

x = 3 — 2

The zero of the function is x = 3 — 2 .


g(x) = 0.

x2 + 22x + 121 = 0 (x + 11)2 = 0 x + 11 = 0 x = −11

The zero of the function is x = −11.


h(x) = 0.

x2 + 19x + 84 = 0 (x + 12)(x + 7) = 0 x + 12 = 0 or x + 7 = 0 x = −12 or x = −7

The zeros of the function are x = −12 and x = −7.

55. Because 8 and 11 are the zeros of the function f, the graph

of f has x-intercepts of 8 and 11. Use the x-intercepts and the

intercept form to write the equation y = (x − 8)(x − 11).

Rewrite the equation as y = x2 − 19x + 88, and then

use the equation to provide the rule for function,

f (x) = x2 − 19x + 88.

56. Sample answer: The numbers 6 and 14 are equidistant

from the number 10 on the number line. Use these as the

x-intercepts. Use the x-intercepts and intercept form to write

the equation y = (x − 6)(x − 14). Rewrite the equation in

standard form y = x2 − 20x + 84.

57. Step 1 Defi ne the variables. Let x represent the price

increase and R(x) represent the daily revenue.

Step 2 Write a verbal model. Then write and simplify a

quadratic equation.

Daily

revenue=

Number of

sandwiches ⋅ Price per

sandwich

R(x) = (330 + 15x)(6 − 0.25x)

Step 3 Identify the zeros and fi nd their average. Then fi nd

how much each sandwich should cost to maximize

the daily revenue.

The zeros of the revenue function are −22 and 24.

The average of the zeros is −22 + 24 —

2 = 1.

To maximize the revenue, each sandwich should cost

$6.00 − $0.25 = $5.75.

Step 4 Find the maximum daily revenue.

R(1) = [ 330 + 15(1) ] [ 6 − 0.25(1) ] = $1983.75.

58. Step 1 Defi ne the variables. Let x represent the price

increase and R(x) represent the daily revenue.

Step 2 Write a verbal model. Then write and simplify a

quadratic equation.

Monthly

revenue=

Number of

pairs of shoes ⋅ Price per

pair of shoe

R(x) = (200 − 2x)(120 + 2x)

R(x) = 4(100 − x)(60 + x)

Step 3 Identify the zeros and fi nd their average. Then

fi nd how much each pair of shoes should cost to

maximize the monthly revenue.

The zeros of the revenue function are 100 and −60.

The average of the zeros is 100 + (−60)

—— 2 = 20.

To maximize the revenue, each pair of shoes should

cost $120 + $40 = $160.

Step 4 Find the maximum daily revenue.

R(20) = 4(100 − 20)(60 + 20) = $25,600

59. a. The initial height is 188, so the model is

h(t) = −16t2 + 188. Find the zeros of the function.

h(t) = −16t2 + 188

0 = −16t2 + 188

−188 = −16t2

47

— 4 = t2

± √—

47

— 4 = t

± √—

47 —

2 = t

± 3.4 ≈ t

Reject the negative solution, −3.4, because times must be

positive. The log will fall for about 3.4 seconds before it

hits the water.

b. Find h(2) and h(3). These represent the heights after

2 seconds and 3 seconds.

h(2) = −16(2)2 + 188 = 124

h(3) = −16(3)2 + 188 = 44

h(2) − h(3) = 124 − 44 = 80

So, the log fell 80 feet between 2 seconds and 3 seconds.


Chapter 3

60. a. First, fi nd the zeros of the function.

h(t) = 196 − 16t2

0 = 196 − 16t2

16t2 = 196

t2 = 196 —

16

t2 = 49 —

4

t = ± √—

49

— 4

t = ± 7 — 2

The zeros of the function are t = − 7 — 2 and t = 7 —

2 . The

negative zero should be rejected in this situation because

time should be positive, while the positive zero is the

time, in seconds, when the rocks hit the ground.

2 40 t

h

40

60

80

100

120

140

160

180

200

200

Time (seconds)

Hei

gh

t (f

eet)

h(t) = 196 − 16t2

b. The domain represents the time the rocks were in the air and

the range represents the height of the rocks while falling.

61. Write an equation that relates the available fabric to the

intended border. In this case, the equation is

10 = 2x(5 + x) + 2x(4 + x). Solve the equation.

10 = 2x(5 + x) + 2x(4 + x)

5 = x(5 + x) + x(4 + x)

5 = 5x + x2 + 4x + x2

5 = 2x2 + 9x

0 = 2x2 + 9x − 5 0 = (2x − 1)(x + 5)

2x − 1 = 0 or x + 5 = 0

x = 1 — 2 or x = −5

Reject the negative solution because lengths are positive. So,

the width of the border should be 1 —

2 feet, or 6 inches.

62. The situation can be modeled by the function

h(t) = −16t2 + 40. To fi nd how long the seashell is in the

air, fi nd the zeros of the function.

h(t) = −16t2 + 40

0 = −16t2 + 40

16t2 = 40

t2 = 5 — 2

t = ± √—

5 —

2

t = ± √—

10 —

2

t ≈ ±1.6

Reject the negative solution because time is positive. So, the

seashell is in the air for about 1.6 seconds.

63. Find the wind speed needed to have waves of height 5 feet

and 20 feet. Let the height be 5 and solve the equation.

5 = 0.019s2

5 —

0.019 = s2

± √— 5 —

0.019 = s

± 16.22 ≈ s Reject the negative solution because speed is positive.

To have a 5-foot wave, the wind speed should be about

16.22 knots. Let the height be 20 and solve the equation.

20 = 0.019s2

20 —

0.019 = s2

± √— 20 —

0.019 = s

±32.44 ≈ s Reject the negative solution because speed is positive. To

have a 20-foot wave, the wind speed should be about

32.44 knots. The 20-foot wave requires a wind speed twice

as great as the wind speed required for a 5-foot wave.

64. Let x be an odd integer. The next consecutive odd integer is

x + 2. Because the product is 143, the equation related to the

product is x(x + 2) = 143. Next, solve the equation.

x(x + 2) = 143

x2 + 2x = 143

x2 + 2x − 143 = 0 (x + 13)(x − 11) = 0 x + 13 = 0 or x − 11 = 0 x = −13 x = 11

For x = −13, x + 2 = −13 + 2 = −11. So, two

consecutive odd integers whose product is 143 are −13

and −11.

For x = 11, x + 2 = 11 + 2 = 13. So, two consecutive odd

integers whose product is 143 are 11 and 13.


Chapter 3

65. First, write an equation that relates the length of the diagonal

to the lengths of the sides of the quadrilateral. By using the

Pythagorean Theorem, the equation is

(6x)2 + (8x)2 = (5x)2 + 3002. Next, solve the equation.

(6x)2 + (8x)2 = (5x)2 + 3002

36x2 + 64x2 = 25x2 + 90,000

100x2 = 25x2 + 90,000

75x2 = 90,000

x2 = 1200

x = ± √—

1200

x = ±20 √—

3

Reject the negative solution because lengths are positive. So,

x = 20 √—

3 ≈ 34.64. So, the lengths of the remaining sides

are:

6x = 6 ( 20 √—

3 ) = 120 √—

3 ≈ 207.85 feet

8x = 8 ( 20 √—

3 ) = 160 √—

3 ≈ 277.13 feet

5x = 5 ( 20 √—

3 ) = 100 √—

3 ≈ 173.21 feet

66. a. Because the graph does not cross the x-axis, a is positive.

b. Because the parabola opens up and the vertex has been

translated down into the fourth quadrant, there are going

to be two x-intercepts.

67. The rock will hit the ground on Jupiter fi rst. Because the fi rst

term is negative, the height of the falling object will decrease

faster as g gets larger.

68. First, write the equation that models the situation using the

area, 329 + 25 ⋅ 15 = (15 + x)(25 + x). Next, solve the

equation.

329 + 25 ⋅ 15 = (15 + x)(25 + x)

329 + 375 = 375 + 40x + x2

329 = x2 + 40x

0 = x2 + 40x − 329

0 = (x + 47)(x − 7)

x + 47 = 0 or x − 7 = 0 x = −47 or x = 7 Reject the negative solution because lengths are positive. The

patio should be extended 7 feet.

69.

6 120 x

y

4

6

8

10

20

Distance (inches)

Hei

gh

t (i

nch

es)

Flea Jump

y = −0.189x2 + 2.462x

First, fi nd the vertex. Find the x-coordinate.

x = − b — 2a

= − 2.462 —

2(−0.189) ≈ 6.5

Now fi nd the y-coordinate of the vertex.

y = −0.189(6.51)2 + 2.462(6.5) ≈ 8.0

So, the vertex is about (6.5, 8.0). The vertex indicates that

the fl ea’s maximum jump is 6.5 inches away from and

8.0 inches above the starting point. Now, fi nd the zeros of the

equation.

y = −0.189x2 + 2.462x

0 = −0.189x2 + 2.462x

0 = x(−0.189x + 2.462)

x = 0 or −0.189x + 2.462 = 0

x = 0 or x = 2.462 —

0.189 ≈ 13.0

The zeros are 0 and about 13.0. The zeros are the locations

where the fl ea is on the ground.

70. a. The initial height of the paint brush is 50 feet.

b. It takes about 1.77 seconds for the paint brush to hit the

ground because this is the point in time when the height is 0.

71.

x

y

20

60

80

4−4−8−12

You are correct. The graph does not cross the x-axis.


Chapter 3

72. x2 − 4 = (x + 2)(x − 2)

x2 − 9 = (x + 3)(x − 3)

The factored form of x2 − a2 is (x + a)(x − a).

y

x

y = x2 − a2

(−a, 0)

(0, −a2)

(a, 0)

x = 0

73. a. You want to fi nd numbers m and n such that

x2 + a2 = (x + m)(x + n). So, expand the right-hand side

of x2 + a2 = (x + m)(x + n) and write two equations

from the rewritten form.

x2 + a2 = (x + m)(x + n)

x2 + a2 = x2 + (m + n)x + mn

The two equation are m + n = 0 and mn = a2.

b. If m + n = 0, then m = −n. Substitute into the second

equation.

mn = a2

(−n)n = a2

−n2 = a2

n2 = −a2

n = ± √—

−a2

n = ±a √—

−1

Because m = −n and because the square of a real number

can never be negative, you can conclude that m and n are

not real numbers.

74. Sample answer:

6 ft

6 + 2x

x

x4 ft4 + 2x

(4 + 2x)(6 + 2x) = 48

24 + 20x + 4x2 = 48

4x2 + 20x − 24 = 0 4(x2 + 5x − 6) = 0 4(x + 6) (x − 1) = 0 x + 6 = 0 or x − 1 = 0 x = −6 or x = 1

Reject the solution x = −6 because lengths are positive.

So, x = 1.

6 + 2(1) = 6 + 2 = 8 ft

4 + 2(1) = 4 + 2 = 6 ft

The dimensions of the new raft are 6 feet by 8 feet.

75. Write an equation that represents the area of the new parking lot.

2 [ (165 + 75)(300 + 75) − 165 ⋅ 300 ] = (x + 75 + 165)

(x + 75 + 300) − 165 ⋅ 300

Next, solve the equation.

2 [ (165 + 75)(300 + 75) − 165 ⋅ 300 ] = (x + 75 + 165)

(x + 75 + 300) − 165 ⋅ 300

2 [ (240)(375) − 49,500 ] = (x + 240)(x + 375) − 49,500

2(240)(375) − 2 ⋅ 49,500 = (x + 240)(x + 375) − 49,500

2(240)(375) − 49,500 = (x + 240)(x + 375)

180,000 − 49,500 = x2 + 615x + 90,000

130,500 = x2 + 615x + 90,000

0 = x2 + 615x − 40,500

0 = (x + 675)(x − 60)

x + 675 = 0 or x − 60 = 0 x = −675 or x = 60

Reject the negative solution because length is positive. So,

the parking lot needs to be extended 60 feet.

Maintaining Mathematical Profi ciency

76. (x2 + 2) + (2x2 − x) = (x2 + 2x2) + 2 − x = 3x2 − x + 2

77. (x3 + x2 − 4) + (3x2 + 10) = x3 + (x2 + 3x2) + (−4 + 10)

= x3 + 4x2 + 6

78. (−2x + 1) − (−3x2 + x) = −2x + 1 + 3x2 − x = 3x2 − 3x + 1

79. (−3x3 + x2 − 12x) − (−6x2 + 3x − 9) = −3x3 + x2 − 12x + 6x2 − 3x + 9

= −3x3 + 7x2 − 15x + 9

80. (x + 2) (x − 2) = x2 − 2x + 2x − 4 = x2 − 4

81. 2x(3 − x + 5x2) = 2x(3) + 2x(−x) + 2x(5x2)

= 10x3 − 2x2 + 6x

82. (7 − x)(x − 1) = 7x − 7 − x2 + x = −x2 + 8x − 7

83. 11x(−4x2 + 3x + 8) = 11x(−4x2) + 11x(3x) + 11x(8)

= −44x3 + 33x2 + 88x


Chapter 3


1. a. √—

9 is in the subsets of real numbers, rational numbers,

integers, whole numbers, and natural numbers.

b. √—


integers, and whole numbers.

c. − √—


and integers.

d. √—

4 —

9 is in the subsets of real numbers and rational numbers.

e. √—

2 is in the subsets of real numbers and irrational numbers.

f. √—

−1 is in the subset of imaginary numbers.

2. a. F; x2 − 4 = 0 x2 = 4 x = ± √

— 4 = ±2

b. A; x2 + 1 = 0

x2 = −1

x = ± √—

−1 = ±i

c. E; x2 − 1 = 0 x2 = 1 x = ± √

— 1 = ±1

d. D; x2 + 4 = 0 x2 = −4

x = ± √—

−4 = ±2i

e. C; x2 − 9 = 0 x2 = 9 x = ± √

— 9 = ±3

f. B; x2 + 9 = 0 x2 = −9

x = ± √—

−9 = ±3i

3. Subset of Complex Numbers ExampleImaginary numbers √

— −2

Real numbers πIrrational numbers √

— 3

Rational numbers 5 —

9

Integers −4

Whole numbers 0

Natural numbers 10

4. It is possible for a number to be whole and natural, or natural

and rational because one is a subset of the other, but it is not

possible for a number to be rational and irrational, or real

and imaginary because they are separate categories.


1. √—

−4 = √—

4 ⋅ √—

−1

= 2i

2. √—

−12 = √—

12 ⋅ √—

−1

= 2i √—

3

3. − √—

−36 = − √—

36 ⋅ √—

−1

= −6i

4. 2 √—

−54 = 2 √—

54 ⋅ √—

−1

= 6i √—

6

5. Set the real parts equal to each other and the imaginary parts

equal to each other.

x = 9 3 = −y

y = −3

So, x = 9 and y = −3.



9 = −2 x 4y = 3

x = − 9 — 2 y = 3 —

4

So, x = − 9 — 2 and y = 3 —

4 .

7. The impedance of the circuit will be 5 + 3i + (−7i) = 5 − 4i.

8. (9 − i) + (−6 + 7i) = (9 − 6) + (−1 + 7)i

= 3 + 6i

9. (3 + 7i) − (8 − 2i) = (3 − 8) + (7 + 2)i

= −5 + 9i

10. −4 − (1 + i) − (5 + 9i) = (−4 − 1 − 5) + (−1 − 9)i

= −10 − 10i

11. (−3i)(10i) = −30i2

= −30(−1)

= 30

12. i(8 − i) = 8i − i2

= 8i − (−1)

= 1 + 8i

13. (3 + i)(5 − i) = 15 − 3i + 5i − i2

= 15 + 2i − (−1)

= 16 + 2i


Chapter 3

14. x2 = −13

x = ± √—

−13

x = ±i √—

13

The solutions are −i √—

13 and i √—

13 .

15. x2 = −38

x = ± √—

−38

x = ±i √—

38


38 and i √—

38 .

16. x2 + 11 = 3 x2 = −8

x = ± √—

−8

x = ±2i √—

2

The solutions are −2i √—

2 and 2i √—

2 .

17. x2 − 8 = −36

x2 = −28

x = ± √—

−28

x = ±2i √—

7


7 and 2i √—

7 .

18. 3x2 − 7 = −31

3x2 = −24

x2 = −8

x = ± √—

−8

x = ±2i √—

2


2 and 2i √—

2 .

19. 5x2 + 33 = 3 5x2 = −30

x2 = −6

x = ± √—

−6

x = ±i √—

6


6 and i √—

6 .

20. x2 + 7 = 0

x2 = −7

x = ± √—

−7

x = ±i √—

7

So, the zeros of f are i √—

7 and −i √—

7 .

21. −x2 − 4 = 0 −x2 = 4 x2 = −4

x = ± √—

−4

x = ±2i

So, the zeros of f are 2i and −2i.

22. 9x2 + 1 = 0 9x2 = −1

x2 = − 1 — 9

x = ± √—

− 1 —

9

x = ± 1 — 3 i

So, the zeros of f are 1 —

3 i and − 1 —

3 i.



1. The imaginary unit is i = √—

−1 and is used to write the

square root of any negative number.

2. For a complex number 5 + 2i, the imaginary part is 2i and

the real part is 5.

3. To add two complex numbers, add the real parts and the

imaginary parts separately.

4. 3 + 0i does not belong because it is the only number that is

not an imaginary number.


5. √—

−36 = √—

36 ⋅ √—

−1

= 6i

6. √—

−64 = √—

64 ⋅ √—

−1

= 8i

7. √—

−18 = √—

18 ⋅ √—

−1

= 3i √—

2

8. √—

−24 = √—

24 ⋅ √—

−1

= 2i √—

6

9. 2 √—

−16 = 2 √—

16 ⋅ √—

−1

= 8i

10. −3 √—

−49 = −3 √—

49 ⋅ √—

−1

= −21i

11. −4 √—

−32 = −4 √—

32 ⋅ √—

−1

= −16i √—

2

12. 6 √—

−63 = 6 √—

63 ⋅ √—

−1

= 18i √—

7



4x = 8 2 = y x = 2 y = 2 So, x = 2 and y = 2.


Chapter 3



3x = 27 6 = y x = 9 y = 6 So, x = 9 and y = 6.



−10x = 20 12 = 3y

x = −2 y = 4 So, x = −2 and y = 4.



9x = −36 −18 = 6y

x = −4 y = −3

So, x = −4 and y = −3.



2x = 14 −y = 12

x = 7 y = −12

So, x = 7 and y = −12.



−12x = 60 y = −13

x = −5

So, x = −5 and y = −13.



54 = 9x − 1 — 7 y = −4

x = 6 y = 28

So, x = 6 and y = 28.



15 = 1 — 2 x −3y = 2

x = 30 y = − 2 — 3

So, x = 30 and y = − 2 — 3 .

21. (6 − i) + (7 + 3i) = (6 + 7) + (−1 + 3)i

= 13 + 2i

22. (9 + 5i) + (11 + 2i) = (9 + 11) + (5 + 2)i

= 20 + 7i

23. (12 + 4i) − (3 − 7i) = (12 − 3) + (4 + 7)i

= 9 + 11i

24. (2 − 15i) − (4 + 5i) = (2 − 4) + (−15 − 5)i

= −2 − 20i

25. (12 − 3i) + (7 + 3i) = (12 + 7) + (−3 − 3)i

= 19

26. (16 − 9i) − (2 − 9i) = (16 − 2) + (−9 + 9)i

= 14

27. 7 − (3 + 4i) + 6i = (7 − 3) + (−4 + 6)i

= 4 + 2i

28. 16 − (2 − 3i) − i = (16 − 2) + (3 − 1)i

= 14 + 2i

29. −10 + (6 − 5i) − 9i = (−10 + 6) + (−5 − 9)i

= −4 − 14i

30. −3 + (8 + 2i) + 7i = (−3 + 8) + (2 + 7)i

= 5 + 9i

31. a. √—

−9 + √—

−4 − √—

16 = 3i + 2i − 4 = −4 + 5i

b. √—

−16 + √—

8 + √—

−36 = 4i + 2 √—

2 + 6i

= 2 √—

2 + 10i

32. a. The additive inverse is za = −1 − i. b. The additive inverse is za = −3 + i. c. The additive inverse is za = 2 − 8i.

33. The resistor has a resistance of 12 ohms, so its impedance

is 12 ohms. The inductor has a reactance of 9 ohms, so its

impedance is 9i ohms. The capacitor has a reactance of

7 ohms, so its impedance is −7i ohms.

Impedance of circuit = 12 + 9i + (−7i) = 12 + 2i

The impedance of the circuit is (12 + 2i) ohms.





Impedance of circuit = 4 + 6i + (−9i) = 4 − 3i

The impedance of the circuit is (4 − 3i) ohms.





Impedance of circuit = 8 + 3i + (−2i) = 8 + i The impedance of the circuit is (8 + i) ohms.


Chapter 3





Impedance of circuit = 14 + 7i + (−8i) = 14 − i The impedance of the circuit is (14 − i) ohms.

37. 3i(−5 + i) = −15i + 3i 2

= −15i + 3(−1)

= −3 − 15i

38. 2i(7 − i) = 14i − 2i 2

= 14i − 2(−1)

= 2 + 14i

39. (3 − 2i) (4 + i) = 12 + 3i − 8i − 2i 2

= 12 − 5i − 2(−1)

= 14 − 5i

40. (7 + 5i) (8 − 6i) = 56 − 42i + 40i − 30i 2

= 56 − 2i − 30(−1)

= 86 − 2i

41. (4 − 2i) (4 + 2i) = 16 + 8i − 8i − 4i 2

= 16 − 4(−1)

= 16 + 4 = 20

42. (9 + 5i) (9 − 5i) = 81 − 45i + 45i − 25i 2

= 81 − 25(−1)

= 81 + 25

= 106

43. (3 − 6i)2 = (3 − 6i)(3 − 6i)

= 9 − 18i − 18i + 36i2

= 9 − 36i + 36(−1)

= −27 − 36i

44. (8 + 3i)2 = (8 + 3i)(8 + 3i)

= 64 + 24i + 24i + 9i2

= 64 + 48i + 9(−1)

= 55 + 48i

45. 11 − (4 + 3i) + 5i = [(11 − 4) − 3i] + 5i Distributive Property

= (7 − 3i) + 5i Simplify.

= 7 + (−3 + 5)i Defi nition of complex addition

= 7 + 2i Write in standard form.

46. (3 + 2i) (7 − 4i) = 21 − 12i + 14i − 8i2 Multiply using FOIL.

= 21 + 2i − 8(−1) Simplify and use i2 = −1.

= 21 + 2i + 8 Simplify.

= 29 + 2i Write in standard form.

47. (6 − 7i) − (4 − 3i) = (6 − 4) + (−7 + 3)i

= 2 − 4i Place the tiles as 6, 7, 4, and 3.

48. 2i(−5 + 9i) = 2i(−5) + 2i(9i)

= −10i + 18i2

= −10i + 18(−1)

= −18 − 10i Place the tiles as 2, −5, and 9.

49. x2 + 9 = 0 x2 = −9

x = ± √—

−9

x = ±3i

The solutions are x = −3i and x = 3i.

50. x2 + 49 = 0 x2 = −49

x = ± √—

−49

x = ±7i

The solutions are x = −7i and x = 7i.

51. x2 − 4 = −11

x2 = −7

x = ± √—

−7

x = ±i √—

7

The solutions are x = −i √—

7 and x = i √—

7 .

52. x2 − 9 = −15

x2 = −6

x = ± √—

−6

x = ±i √—

6

The solutions are x = −i √—

6 and x = i √—

6 .

53. 2x2 + 6 = −34

2x2 = −40

x2 = −20

x = ± √—

−20

x = ±2i √—

5

The solutions are x = −2i √—

5 and x = 2i √—

5 .


Chapter 3

54. x2 + 7 = −47

x2 = −54

x = ± √—

−54

x = ±3i √—

6

The solutions are x = −3i √—

6 and x = 3i √—

6.

55. 3x2 + 6 = 0 3x2 = −6

x2 = −2

x = ± √—

−2

x = ±i √—

2

So, the zeros of f are i √—

2 and − i √—

2 .

56. 7x2 + 21 = 0 7x2 = −21

x2 = −3

x = ± √—

−3

x = ±i √—

3

So, the zeros of g are i √—

3 and −i √—

3 .

57. 2x2 + 72 = 0 2x2 = −72

x2 = −36

x = ± √—

−36

x = ±6i

So, the zeros of h are 6i and −6i.

58. −5x2 − 125 = 0 −5x2 = 125

x2 = −25

x = ± √—

−25

x = ±5i

So, the zeros of k are 5i and −5i.

59. −x2 − 27 = 0 −x2 = 27

x2 = −27

x = ± √—

−27

x = ±3i √—

3

So, the zeros of m are 3i √—

3 and −3i √—

3 .

60. x2 + 98 = 0 x2 = −98

x = ± √—

−98

x = ±7i √—

2

So, the zeros of p are 7i √—

2 and −7i √—

2 .

61. − 1 — 2 x2 − 24 = 0

− 1 — 2 x2 = 24

x2 = −48

x = ± √—

−48

x = ±4i √—

3

So, the zeros of r are 4i √—

3 and −4i √—

3 .

62. − 1 — 5 x2 − 10 = 0

− 1 — 5 x2 = 10

x2 = −50

x = ± √—

−50

x = ±5i √—

2

So, the zeros of f are 5i √—

2 and −5i √—

2 .

63. i2 was not simplifi ed.

(3 + 2i)(5 − i) = 15 − 3i + 10i − 2i2

= 15 + 7i − 2(−1)

= 17 + 7i

64. Squaring a complex number requires FOIL.

(4 + 6i)2 = (4 + 6i)(4 + 6i)

= 16 + 24i + 24i + 36i2

= 16 + 48i + 36(−1)

= −20 + 48i

65. a. (−4 + 7i) + (−4 − 7i) = −8

b. (2 − 6i) − (−10 + 4i) = 12 − 10i

c. (25 + 15i) − (25 − 6i) = 21i

d. (5 + i) (8 − i) = 40 − 5i + 8i − i2 = 41 + 3i

e. (17 − 3i) + (−17 − 6i) = −9i

f. (−1 + 2i)(11 − i) = −11 + i + 22i − 2i2 = −9 + 23i

g. (7 + 5i) + (7 − 5i) = 14

h. (−3 + 6i) − (−3 − 8i) = 14i

Real numbers

Imaginary numbers

Pure imaginary numbers

−8

14

12 − 10i41 + 3i

−9 + 23i

21i −9i14i


Chapter 3

66. Your friend is incorrect. By using the defi nition of the

imaginary number, you have

√—

−4 ⋅ √—

−9 = 2i ⋅ 3i

= 6i2

= −6.

67. Powers of i Simplifi ed formi i

i2 −1

i3 −ii4 1

i5 i

i6 −1

i7 −ii8 1

i9 i

i10 −1

i11 −ii12 1

The results of in alternate in the pattern i, −1, −i, and 1.

68. The functions f and g both have real zeros because they both

have x-intercepts, and h has imaginary zeros because it does

not have any x-intercepts.

69. (3 + 4i) − (7 − 5i) + 2i(9 + 12i) = 3 + 4i − 7 + 5i + 18i + 24i2

= −4 + 27i + 24(−1)

= −28 + 27i

70. 3i(2 + 5i) + (6 − 7i) − (9 + i) = 6i + 15i2 + 6 − 7i − 9 − i = −3 − 2i + 15(−1)

= −18 − 2i

71. (3 + 5i)(2 − 7i4) = (3 + 5i) [ 2 − 7(1) ]

= (3 + 5i)(−5)

= −15 − 25i

72. 2i3(5 − 12i) = 2(−i)(5 − 12i)

= −10i + 24i2

= −10i + 24(−1)

= −24 − 10i

73. (2 + 4i5) + (1 − 9i6) − (3 + i7) = (2 + 4i) + [ 1 − 9(−1) ] − [ 3 + (−i) ]

= (2 + 4i) + 10 − (3 − i) = 9 + 5i

74. (8 − 2i4) + (3 − 7i8) − (4 + i9) = [8 − 2(1)] + [3 − 7(1)] − (4 + i)

= 6 − 4 − (4 + i) = −2 − i

75. Sample answer: The imaginary numbers 1 + i and 1 − i have a sum of 1 + i + 1 − i = 2 and a product of

(1 + i)(1 − i) = 1 − i + i − (i)2 = 2. The real parts are

equal and the imaginary parts are opposites.

76. Method 1 distributes 4i to each term, then simplifi es.

Method 2 factors 4i out of each term, combines like terms,

and simplifi es. Sample answer: Method 1 is preferred

because it requires fewer steps.

77. a. The statement is false. Counterexample:

(1 + i) + (1 − i) = 2 b. The statement is true. Example: (3i)(2i) = 6i2 = −6

c. The statement is true. Example: 2i = 0 + 2i

d. The statement is false. Counterexample: 1 + i is a

complex number but not a real number.

78. Sample answer:

14Ω

2Ω 5Ω


79. 3(x − 2) + 4x − 1 = x − 1

3(1 − 2) + 4(1) − 1 =?

(1) − 1

3(−1) + 4 − 1 =?

0

−3 + 3 =?

0

0 = 0 ✓

This is a valid equation, so x = 1 is a solution of the equation.

80. x3 − 6 = 2x2 + 9 − 3x

(−5)3 − 6 =?

2(−5)2 + 9 − 3(−5)

−125 − 6 =?

2(25) + 9 + 15

−131 =?

50 + 21

−131 ≠ 71 ✗

This is not a valid equation, so x = −5 is not a solution of

the equation.

81. −x2 + 4x = 19 —

3 x2

− ( − 3 — 4 )

2

+ 4 ( − 3 — 4 ) =? 19

— 3 ( − 3 —

4 )

2

− 9 — 16

− 3 =?

19 —

3 ( 9 —

16 )

− 57 —

16 =

? 171

— 48

− 57 —

16 ≠ 57 —

16 ✗

This is not a valid equation, so x = − 3 — 4 is not a solution of

the equation.


Chapter 3

82. Use the vertex to fi nd a in the vertex form.

y = a(x − h)2 + k 3 = a(0 − 1)2 + 2 1 = a So, the equation for the parabola is y = (x − 1)2 + 2.


y = a(x − h)2 + k 5 = a(−1 + 3)2 − 3 8 = 4a

2 = a So, the equation for the parabola is y = 2(x + 3)2 − 3.


y = a(x − h)2 + k −2 = a(3 − 2)2 − 1 −1 = a So, the equation for the parabola is y = −(x − 2)2 − 1.

3.3 Explorations (p.111)

1. a.

b. You will need nine 1-tiles to complete the square.

c. The value for c is 9.

d. x2 + 6x + 9 = (x + 3)2

2. a.

Expression

Value of c needed to complete the

square

Expression written as a

binomial squared

x2 + 2x + c 1 (x + 1)2

x2 + 4x + c 4 (x + 2)2

x2 + 8x + c 16 (x + 4)2

x2 + 10x + c 25 (x + 5)2

b. The relationships are that d = b — 2 and c = d2 or d = √—

c .

c. The number in the second column is found by fi nding half

of b, then squaring it, or ( b — 2 )

2

.

3. To complete a square for a quadratic expression, add b2

— 4 to

the expression x2 + bx.

4. To solve the equation x2 + 6x = 1, add b2

— 4 = 6

2

— 4 = 9 to

each side of the equation and rewrite the equation as

(x + 3)2 = 10. Then use the square root method to solve the

rewritten equation.

3.3 Monitoring Progress (pp.112–115)

1. x2 + 4x + 4 = 36

(x + 2)2 = 36

x + 2 = ±6

x = −2 ± 6 So, the solutions are x = −8 and x = 4.

2. x2 − 6x + 9 = 1 (x − 3)2 = 1 x − 3 = ±1

x = 3 ± 1 So, the solutions are x = 2 and x = 4.

3. x2 − 22x + 121 = 81

(x − 11)2 = 81

x − 11 = ± 9 x = 11 ± 9 So, the solutions are x = 2 and x = 20.

4. Step 1 Find half the coeffi cient of x. 8 —

2 = 4

Step 2 Square the result of Step 1. 42 = 16

Step 3 Replace c with the result of Step 2. x2 + 8x + 16

The expression x2 + 8x + c is a perfect square trinomial

when c = 16. Then

x2 + 8x + 16 = (x + 4)(x + 4) = (x + 4)2.

5. Step 1 Find half the coeffi cient of x. −2

— 2 = −1

Step 2 Square the result of Step 1. (−1)2 = 1 Step 3 Replace c with the result of Step 2. x2 − 2x + 1 The expression x2 − 2x + c is a perfect square trinomial

when c = 1. Then

x2 − 2x + 1 = (x − 1) (x − 1) = (x − 1)2.


— 2 = − 9 —

2

Step 2 Square the result of Step 1. ( − 9 — 2 )

2

= 81 —

4

Step 3 Replace c with the result of Step 2. x2 − 9x + 81 —

4

The expression x2 − 9x + c is a perfect square trinomial

when c = 81 —

4 . Then

x2 − 9x + 81 —

4 = ( x − 9 —

2 ) ( x − 9 —

2 ) = ( x − 9 —

2 )

2

.


Chapter 3

7. x2 − 4x + 8 = 0 x2 − 4x = −8

x2 − 4x + 4 = −8 + 4 (x − 2)2 = −4

x − 2 = ± √—

−4

x − 2 = ±2i

x = 2 ± 2i

The solutions are x = 2 − 2i and x = 2 + 2i.

8. x2 + 8x − 5 = 0 x2 + 8x = 5

x2 + 8x + 16 = 5 + 16

(x + 4)2 = 21

x + 4 = ± √—

21

x = −4 ± √—

21


21 and x = −4 + √—

21 .

9. −3x2 − 18x − 6 = 0

x2 + 6x + 2 = 0 x2 + 6x = −2

x2 + 6x + 9 = −2 + 9 (x + 3)2 = 7 x + 3 = ± √

— 7

x = −3 ± √—

7


7 and x = −3 + √—

7 .

10. 4x2 + 32x = −68

x2 + 8x = −17

x2 + 8x + 16 = −17 + 16

(x + 4)2 = −1

x + 4 = ± √—

−1

x = −4 ± i

The solutions are x = −4 − i and x = −4 + i.

11. 6x(x + 2) = −42

x(x + 2) = −7

x2 + 2x = −7

x2 + 2x + 1 = −7 + 1 (x + 1)2 = −6

x + 1 = ± √—

−6

x = −1 ± i √—

6

The solutions are x = −1 − i √—

6 and x = −1 + i √—

6 .

12. 2x(x − 2) = 200

x(x − 2) = 100

x2 − 2x = 100

x2 − 2x + 1 = 100 + 1 (x − 1)2 = 101

x − 1 = ± √—

101

x = 1 ± √—

101


101 and x = 1 + √—

101 .

13. y = x2 − 8x + 18

y + ? = (x2 − 8x + ?) + 18

y + 16 = (x2 − 8x + 16) + 18

y + 16 = (x − 4)2 + 18

y = (x − 4)2 + 2 The vertex form of the function is y = (x − 4)2 + 2. The

vertex is (4, 2).

14. y = x2 + 6x + 4 y + ? = (x2 + 6x + ?) + 4 y + 9 = (x2 + 6x + 9) + 4 y + 9 = (x + 3)2 + 4 y = (x + 3)2 − 5 The vertex form of the function is y = (x + 3)2 − 5. The

vertex is (−3, −5).

15. y = x2 − 2x − 6 y + ? = (x2 − 2x + ?) − 6

y + 1 = (x2 − 2x + 1) − 6 y + 1 = (x − 1)2 − 6 y = (x − 1)2 − 7 The vertex form of the function is y = (x − 1)2 − 7. The

vertex is (1, −7).

16. Write the function in vertex form by completing the square.

y = −16t2 + 80t + 2 y = −16(t2 − 5t) + 2 y + ? = −16(t2 − 5t + ?) + 2

y + (−16) ( 25 —

4 ) = −16 ( t2 − 5t + 25

— 4 ) + 2

y − 100 = −16 ( t − 5 — 2 )

2

+ 2

y = −16 ( t − 5 — 2 )

2

+ 102

The vertex is ( 5 — 2 , 102 ) . Find the zeros of the function.

0 = −16(t − 2.5)2 + 102

−102 = −16(t − 2.5)2

6.375 = (t − 2.5)2

± √—

6.375 = t − 2.5

2.5 ± √—

6.375 = t Reject the negative solution t = 2.5 − √

— 6.375 ≈ −0.02

because time must be positive. So, the maximum height of

the ball is 102 feet, and it takes 2.5 + √—

6.375 ≈ 5 seconds

for the ball to hit the ground.


Chapter 3



1. To complete the square for the expression x2 + bx, add ( b — 2 )

2

.

2. The trinomial x2 − 6x + 9 is a perfect square trinomial

because it equals (x − 3)2.


3. x2 − 8x + 16 = 25

(x − 4)2 = 25

x − 4 = ±5

x = 4 ± 5 The solutions are x = −1 and x = 9.

4. r2 − 10r + 25 = 1 (r − 5)2 = 1 r − 5 = ±1

r = 5 ± 1 The solutions are r = 4 and r = 6.

5. x2 − 18x + 81 = 5 (x − 9)2 = 5 x − 9 = ± √

— 5

x = 9 ± √—

5


5 and x = 9 + √—

5 .

6. m2 + 8m + 16 = 45

(m + 4)2 = 45

m + 4 = ±3 √—

5

m = −4 ± 3 √—

5

The solutions are m = −4 − 3 √—

5 and m = −4 + 3 √—

5 .

7. y2 − 24y + 144 = −100

(y − 12)2 = −100

y − 12 = ±10i

y = 12 ± 10i The solutions are y = 12 − 10i and y = 12 + 10i.

8. x2 − 26x + 169 = −13

(x − 13)2 = −13

x − 13 = ±i √—

13

x = 13 ± i √—

13

The solutions are x = 13 − i √—

13 and x = 13 + i √—

13 .

9. 4w2 + 4w + 1 = 75

(2w + 1)2 = 75

2w + 1 = ±5 √—

3

2w = −1 ± 5 √—

3

w = −1 ± 5 √—

3 —

2

The solutions are w = −1 − 5 √—

3 —

2 and w = −1 + 5 √

— 3 —

2 .

10. 4x2 − 8x + 4 = 1 4(x2 − 2x + 1) = 1 4(x − 1)2 = 1

(x − 1)2 = 1 — 4

x − 1 = ± 1 — 2

x = 1 ± 1 — 2

The solutions are x = 1 — 2 and x = 3 —

2 .

11. Step 1 Find half the coeffi cient of x. 10

— 2 = 5

Step 2 Square the result of Step 1. (5)2 = 25



when c = 25. Then

x2 + 10x + 25 = (x + 5)(x + 5) = (x + 5)2.


— 2 = 10




when c = 100. Then

x2 + 20x + 100 = (x + 10)(x + 10) = (x + 10)2.

13. Step 1 Find half the coeffi cient of y. −12

— 2 = −6

Step 2 Square the result of Step 1. (−6)2 = 36

Step 3 Replace c with the result of Step 2. y2 − 12y + 36

The expression y2 − 12y + c is a perfect square trinomial

when c = 36. Then

y2 − 12y + 36 = (y − 6)(y − 6) = (y − 6)2 .

14. Step 1 Find half the coeffi cient of t. −22

— 2 = −11


Step 3 Replace c with the result of Step 2. t2 − 22t + 121

The expression t2 − 22t + c is a perfect square trinomial

when c = 121. Then

t2 − 22t + 121 = (t − 11)(t − 11) = (t − 11)2 .


— 2 = −3

Step 2 Square the result of Step 1. (−3)2 = 9 Step 3 Replace c with the result of Step 2. x2 − 6x + 9 The expression x2 − 6x + c is a perfect square trinomial

when c = 9. Then x2 − 6x + 9 = (x − 3)(x − 3) = (x − 3)2.


— 2 = 12




when c = 144. Then

x2 + 24x + 144 = (x + 12)(x + 12) = (x + 12)2.


Chapter 3

17. Step 1 Find half the coeffi cient of z. −5

— 2 = −

5 —

2

Step 2 Square the result of Step 1. ( − 5 — 2 )

2

= 25 —

4

Step 3 Replace c with the result of Step 2. z2 − 5z + 25 —

4

The expression z2 − 5z + c is a perfect square trinomial

when c = 25 —

4 . Then

z2 − 5z + 25 —

4 = ( z − 5 —

2 ) ( z − 5 —

2 ) = ( z − 5 —

2 )

2

.

18. Step 1 Find half the coeffi cient of x. 9 —

2

Step 2 Square the result of Step 1. ( 9 — 2 )

2

= 81 —

4

Step 3 Replace c with the result of Step 2. x2 + 9x + 81 —

4


when c = 81 —

4 . Then

x2 + 9x + 81 —

4 = ( x + 9 —

2 ) ( x + 9 —

2 ) = ( x + 9 —

2 )

2

.

19. Step 1 Find half the coeffi cient of w. 13

— 2

Step 2 Square the result of Step 1. ( 13 —

2 )

2

= 169 —

4

Step 3 Replace c with the result of Step 2. w2 + 13w + 169 —

4

The expression w2 + 13w + c is a perfect square trinomial

when c = 169 —

4 . Then

w2 + 13w + 169 —

4 = ( w + 13

— 2 ) ( w + 13

— 2 ) = ( w + 13

— 2 )

2

.

20. Step 1 Find half the coeffi cient of s. −26

— 2 = −13


Step 3 Replace c with the result of Step 2. s2 − 26s + 169

The expression s2 − 26s + c is a perfect square trinomial

when c = 169. Then

s2 − 26s + 169 = (s − 13)(s − 13) = (s − 13)2.

21. The value of c is 2 ⋅ 2 = 4; x2 + 4x + 4 = (x + 2)2.




25. x2 + 6x + 3 = 0 x2 + 6x = −3

x2 + 6x + 9 = −3 + 9 (x + 3)2 = 6 x + 3 = ± √

— 6

x = −3 ± √—

6


6 and x = −3 + √—

6 .

26. s2 + 2s − 6 = 0 s2 + 2s = 6 s2 + 2s + 1 = 6 + 1 (s + 1)2 = 7 s + 1 = ± √

— 7

s = −1 ± √—

7

The solutions are s = −1 − √—

7 and s = −1 + √—

7 .

27. x2 + 4x − 2 = 0 x2 + 4x = 2 x2 + 4x + 4 = 2 + 4 (x + 2)2 = 6 x + 2 = ± √

— 6

x = −2 ± √—

6


6 and x = −2 + √—

6 .

28. t2 − 8t − 5 = 0 t2 − 8t = 5 t2 − 8t + 16 = 5 + 16

(t − 4)2 = 21

t − 4 = ± √—

21

t = 4 ± √—

21


21 and x = 4 + √—

21 .

29. z(z + 9) = 1 z2 + 9z = 1

z2 + 9z + 81 —

4 = 1 + 81

— 4

( z + 9 — 2 )

2

= 85 —

4

z + 9 — 2 = ± √

— 85 —

2

z = −9 ± √—

85 —

2

The solutions are z = −9 − √—

85 —

2 and z = −9 + √

— 85 —

2 .

30. x(x + 8) = −20

x2 + 8x = −20

x2 + 8x + 16 = −20 + 16

(x + 4)2 = −4

x + 4 = ±2i

x = −4 ± 2i

The solutions are x = −4 − 2i and x = −4 + 2i.


Chapter 3

31. 7t2 + 28t + 56 = 0 t2 + 4t + 8 = 0 t2 + 4t = −8

t2 + 4t + 4 = −8 + 4 (t + 2)2 = −4

t + 2 = ±2i

t = −2 ± 2i

The solutions are t = −2 − 2i and t = −2 + 2i.

32. 6r2 + 6r + 12 = 0 r2 + r + 2 = 0 r2 + r = −2

r2 + r + 1 — 4 = −2 + 1 —

4

( r + 1 — 2 )

2

= − 7 — 4

r + 1 — 2 = ±i

√—

7 —

2

r = −1 ± i √—

7 —

2

The solutions are r = −1 − i √—

7 —

2 and r = −1 + i √

— 7 —

2 .

33. 5x(x + 6) = −50

x(x + 6) = −10

x2 + 6x = −10

x2 + 6x + 9 = −10 + 9 (x + 3)2 = −1

x + 3 = ±i

x = −3 ± i The solutions are x = −3 − i and x = −3 + i.

34. 4w(w − 3) = 24

w(w − 3) = 6 w 2 − 3w = 6

w2 − 3w + 9 — 4 = 6 + 9 —

4

( w − 3 — 2 )

2

= 33 —

4

w − 3 — 2 = ± √

— 33 —

2

w = 3 ± √—

33 —

2

The solutions are w = 3 − √—

33 —

2 and w = 3 − √

— 33 —

2 .

35. 4x2 − 30x = 12 + 10x

4x2 − 40x = 12

x2 − 10x = 3 x2 − 10x + 25 = 3 + 25

(x − 5)2 = 28

x − 5 = ±2 √—

7

x = 5 ± 2 √—

7

The solutions are x = 5 − 2 √—

7 and x = 5 + 2 √—

7 .

36. 3s2 + 8s = 2s − 9 3s2 + 6s = −9

s2 + 2s = −3

s2 + 2s + 1 = −3 + 1 (s + 1)2 = −2

s + 1 = ±i √—

2

s = −1 ± i √—

2

The solutions are s = −1 − i √—

2 and s = −1 + i √—

2 .

37. The constant 4(9) = 36 should have been added to the right

side of the equation instead of 9.

4x2 + 24x − 11 = 0 4(x2 + 6x) = 11

4(x2 + 6x + 9) = 11 + 4(9)

4(x + 3)2 = 47

(x + 3)2 = 47 —

4

x + 3 = ± √—

47 —

2

x = −3 ± √—

47 —

2

38. The number b —

2 was not squared before being added.

x2 + 30x + c

x2 + 30x + ( 30 —

2 )

2

x2 + 30x + 225

39. yes; All of the steps would be the same as with two real

solutions, with the exception of the constant being negative

when you take the square root.

40. The solutions are E and F.

x2 − 2ax + a2 = b2

(x − a)2 = b2

x − a = ±b

x = a ± b

41. Use factoring because the left-hand side factors.

x2 − 4x − 21 = 0 (x − 7)(x + 3) = 0 x − 7 = 0 or x + 3 = 0 x = 7 or x = −3



Chapter 3


x2 + 13x + 22 = 0 (x + 11)(x + 2) = 0 x + 11 = 0 or x + 2 = 0 x = −11 or x = −2


43. Use square roots because the equation is of the form u2 = d.

(x + 4)2 = 16

x + 4 = ±4

x = −4 ± 4

The solutions are x = −8 and x = 0.

44. Use square roots because the equation is of the form u2 = d.

(x − 7)2 = 9 x − 7 = ±3

x = 7 ± 3 The solutions are x = 4 and x = 10.


x2 + 12x + 36 = 0 (x + 6)2 = 0 x + 6 = 0 x = −6

The solution is x = −6.


x2 − 16x + 64 = 0 (x − 8)2 = 0 x − 8 = 0

x = 8 The solution is x = 8.

47. Use completing the square because the equation cannot be

factored or written in the form u2 = d.

2x2 + 4x − 3 = 0 2x2 + 4x = 3 2(x2 + 2x) = 3 2(x2 + 2x + 1) = 3 + 2 2(x + 1)2 = 5

(x + 1)2 = 5 — 2

x + 1 = ± √—

10 —

2

x = −1 ± √—

10 —

2


10 —

2 and x = −1 + √

— 10 —

2 .

48. Use completing the square because the equation cannot be

factored or written in the form u2 = d.

3x2 + 12x + 1 = 0 3x2 + 12x = −1

3(x2 + 4x) = −1

3(x2 + 4x + 4) = −1 + 12

3(x + 2)2 = 11

(x + 2)2 = 11 —

3

x + 2 = ± √—

33 —

3

x = −2 ± √—

33 —

3


33 —

3 and x = −2 + √

— 33 —

3 .

49. Use square roots because the equation can be written in the

form u2 = d.

x2 − 100 = 0 x2 = 100

x = ±10


50. Use square roots because the equation can be written in the

form u2 = d.

4x2 − 20 = 0 4x2 = 20

x2 = 5 x = ± √

— 5


5 and x = √—

5 .

51. The area of the rectangle can be modeled by the equation

50 = x(x + 10). Solve the equation.

50 = x(x + 10)

50 = x2 + 10x

50 + 25 = x2 + 10x + 25

75 = (x + 5)2

±5 √—

3 = x + 5 x = −5 ± 5 √

— 3

Reject the negative solution, −5 − 5 √—

3 , because lengths are

positive. So, x = −5 + 5 √—

3 .

52. The area of the parallelogram can be modeled by the

equation 48 = x(x + 6). Solve the equation.

48 = x (x + 6)

48 = x2 + 6x

48 + 9 = x2 + 6x + 9 57 = (x + 3)2

± √—

57 = x + 3 x = −3 ± √

— 57

Reject the negative solution, −3 − √—


positive. So, x = −3 + √—

57 .


Chapter 3

53. The area of the triangle can be modeled by the equation

40 = 1 — 2 x (x + 4). Solve the equation.

40 = 1 — 2 x (x + 4)

80 = x (x + 4)

80 = x2 + 4x

80 + 4 = x2 + 4x + 4 84 = (x + 2)2

±2 √—

21 = x + 2 x = −2 ± 2 √

— 21

Reject the negative solution, −2 − 2 √—

21 , because lengths

are positive. So, x = −2 + 2 √—

21 .

54. The area of the trapezoid can be modeled by the equation

20 = 1 — 2 x [ (3x − 1) + (x + 9) ] . Solve the equation.

20 = 1 — 2 x [ (3x − 1) + (x + 9) ]

40 = x(4x + 8)

40 = 4x2 + 8x

10 = x2 + 2x

10 + 1 = x2 + 2x + 1 11 = (x + 1)2

± √—

11 = x + 1 x = −1 ± √

— 11

Reject the negative solution, −1 − √—


positive. So, x = −1 + √—

11 .

55. f (x) = x2 − 8x + 19

f (x) + ? = (x2 − 8x + ?) + 19

f (x) + 16 = (x2 − 8x + 16) + 19

f (x) + 16 = (x − 4)2 + 19

f (x) = (x − 4)2 + 3 The vertex form of the function is f (x) = (x − 4)2 + 3.

The vertex is (4, 3).

56. g(x) = x2 − 4x − 1 g(x) + ? = (x2 − 4x + ?) − 1 g(x) + 4 = (x2 − 4x + 4) − 1 g(x) + 4 = (x − 2)2 − 1 g(x) = (x − 2)2 − 5 The vertex form of the function is g(x) = (x − 2)2 − 5.

The vertex is (2, −5).

57. g(x) = x2 + 12x + 37

g(x) + ? = (x2 + 12x + ?) + 37

g(x) + 36 = (x2 + 12x + 36) + 37

g(x) + 36 = (x + 6)2 + 37

g(x) = (x + 6)2 + 1 The vertex form of the function is g(x) = (x + 6)2 + 1.

The vertex is (−6, 1).

58. h(x) = x2 + 20x + 90

h(x) + ? = (x2 + 20x + ?) + 90

h(x) + 100 = (x2 + 20x + 100) + 90

h(x) + 100 = (x + 10)2 + 90

h(x) = (x + 10)2 − 10

The vertex form of the function is h(x) = (x + 10)2 − 10.

The vertex is (−10, −10).

59. h(x) = x2 + 2x − 48

h(x) + ? = (x2 + 2x + ?) − 48

h(x) + 1 = (x2 + 2x + 1) − 48

h(x) + 1 = (x + 1)2 − 48

h(x) = (x + 1)2 − 49

The vertex form of the function is h(x) = (x + 1)2 − 49. The

vertex is (−1, −49).

60. f (x) = x2 + 6x − 16

f (x) + ? = (x2 + 6x + ?) − 16

f (x) + 9 = (x2 + 6x + 9) − 16

f (x) + 9 = (x + 3)2 − 16

f (x) = (x + 3)2 − 25

The vertex form of the function is f (x) = (x + 3)2 − 25.

The vertex is (−3, −25).

61. f (x) = x2 − 3x + 4 f (x) + ? = (x2 − 3x + ?) + 4

f (x) + 9 — 4 = ( x2 − 3x + 9 —

4 ) + 4

f (x) + 9 — 4 = ( x − 3 —

2 )

2

+ 4

f (x) = ( x − 3 — 2 )

2

+ 7 — 4

The vertex form of the function is f (x) = ( x − 3 —

2 )

2

+ 7 — 4 .

The vertex is ( 3 — 2 ,

7 —

4 ) .

62. g(x) = x2 + 7x + 2 g(x) + ? = (x2 + 7x + ?) + 2

g(x) + 49 —

4 = ( x2 + 7x + 49

— 4 ) + 2

g(x) + 49 —

4 = ( x + 7 —

2 ) 2 + 2

g(x) = ( x + 7 — 2 ) 2 − 41

— 4

The vertex form of the function is g(x) = ( x + 7 — 2 ) 2 − 41

— 4 .

The vertex is ( − 7 — 2 , − 41

— 4 ) .


Chapter 3

63. a. Write the function in vertex form by completing the square.

h = −16t2 + 32t + 6 h = −16(t2 − 2t) + 6 h + ? = −16(t2 − 2t + ?) + 6 h + (−16)(1) = −16(t2 − 2t + 1) + 6 h − 16 = −16(t − 1)2 + 6 h = −16(t − 1)2 + 22

The vertex is (1, 22). So, the maximum height of the baton

is 22 feet.

b. Find the time when the height is 4 feet.

4 = −16(t − 1)2 + 22

−18 = −16(t − 1)2

1.125 = (t − 1)2

± √—

1.125 = t − 1 1 ± √

— 1.125 = t

Reject the negative solution, 1 − √—

1.125 ≈ −0.1 because

time must be positive. So, the baton is in the air for

1 + √—

1.125 ≈ 2.1 seconds before it is caught.


h = − 500 —

9 t2 + 1000

— 3 t + 10

h = − 500 —

9 (t2 − 6t) + 10

h + ? = − 500 —

9 (t2 − 6t + ?) + 10

h + ( − 500 —

9 ) (9) = − 500

— 9 (t2 − 6t + 9) + 10

h − 500 = − 500 —

9 (t − 3)2 + 10

h = − 500 —

9 (t − 3)2 + 510

The vertex is (3, 510). So, the fi reworks explode at 510 feet

after being in the air for 3 seconds.

65. a. Use the x-intercepts, 70 and −50, to fi nd the vertex.

x = p + q —

2 = 70 + (−50)

— 2 = 10

y = (70 − 10)(50 + 10) = 3600

The vertex is (10, 3600). So, the maximum weekly

revenue is $3600.

b. Rewrite the equation in vertex form.

y = (70 − x)(50 + x)

y = 3500 + 20x − x2

y = −x2 + 20x + 3500

y = −(x2 − 20x) + 3500

y + ? = −(x2 − 20x + ?) + 3500

y + (−1)(100) = −(x2 − 20x + 100) + 3500

y − 100 = −(x − 10)2 + 3500

y = −(x − 10)2 + 3600

The vertex is (10, 3600). So, the maximum weekly

revenue is $3600.

c. Sample answer: The method in part (a) is preferred

because it requires fewer steps and the vertex of the graph

gives the maximum value.

66. To fi nd the value of h, substitute the value from the point

(0, 9) into the function and solve for h. Then use the negative

solution because the vertex is to the left of the y-axis.

f (x) = (x − h)2

9 = (0 − h)2

9 = h2

±3 = h So, the value of h is −3.

67. Sample answer: Three different methods that can be used to

fi nd the maximum height are:

1. Use the x-intercepts of the graph of the function and then

use their average to fi nd the x-coordinate of the vertex.

Follow by fi nding the y-coordinate of the vertex, which is

the maximum height.

2. Rewrite the function in vertex form and then use the

vertex to fi nd the maximum height.

3. Use the coeffi cients from the original function and

calculate the x-coordinate of the vertex and then the

y-coordinate of the vertex.

Use the coeffi cients of the function to fi nd the x-coordinate

of the vertex.

x = − b — 2a

= − 89.6 —

2(−16) = 2.8

Find the y-coordinate of the vertex.

h = −16(2.8)2 + 89.6(2.8) = 125.44

The vertex of the parabola is (2.8, 125.44). So, the maximum

height of the water is 125.44 feet.


Chapter 3

68. a. The equation that represents the area of the pen is

1512 = x(120 − 2x).

1512 = x(120 − 2x)

1512 = 120x − 2x2

756 = 60x − x2

b. −756 = x2 − 60x

−756 + 900 = x2 − 60x + 900

144 = (x − 30)2

±12 = x − 30

30 ± 12 = x Reject the solution x = 18 because it is too short. So,

x = 42. Thus, the dimensions of the pen are 36 feet by

42 feet.

69. Your friend is incorrect because x2 + 10x + 20 does not

factor into rational numbers.

70. Sample answer: Factor 2 out of the function for f to obtain

f (x) = 2(x2 + 4x + 1). So, the graph of g(x) = x2 + 4x + 1

has the same x-intercepts as f.

First, fi nd the zeros of f.

0 = 2x2 + 8x + 2 0 = x2 + 4x + 1 −1 = x2 + 4x

−1 + 4 = x2 + 4x + 4 3 = (x + 2)2

± √—

3 = x + 2 −2 ± √

— 3 = x

The zeros of f are −2 − √—

3 and −2 + √—

3 .

Next, fi nd the zeros of g.

0 = x2 + 4x + 1 −1 = x2 + 4x

−1 + 4 = x2 + 4x + 4 3 = (x + 2)2

± √—

3 = x + 2 −2 ± √

— 3 = x

The zeros of g are −2 − √—

3 and −2 + √—

3 .

y

2−4 −2−6

−6

2

4

6

x

f(x) = 2x2 + 8x + 2

g(x) = x2 + 4x + 1

71. x2 + bx + c = 0 x2 + bx = −c

x2 + bx + b2

— 4 = −c + b

2

— 4

( x + b — 2 )

2

= b2 − 4c

— 4

x + b — 2 = ± √

— b2 − 4c —

2

x = − b — 2 ± √

— b2 − 4c —

2

x = −b ± √—

b2 − 4c ——

2

72. a. y

2−4 −2

−2

2

4

6

x

y = x2 + 2x

y = (x + 1)2

y

2 4 8−2

−2

−4

−6

−8

6

x

y = x2 − 6x

y = (x − 3)2

b. When completing the square on y = x2 + bx, an

additional term of − b2

— 4 is added in. So, the graph is shifted

vertically, but the axis of symmetry does not change.


Chapter 3

73. The volume of the larger cylinder is given by the expression

9(3 + x)2 π and the volume of the empty space inside of the

pencil holder is 9(9 − x) π. Thus, the volume of clay needed

is V = 9(3 + x)2 π − 9(9 − x) π . Because there are

200 cubic centimeters of clay, solve the equation

200 = 9(3 + x)2 π − 9(9 − x) π to fi nd the thickness of

the pencil holder.

200 = 9(3 + x)2 π − 9(9 − x)π 7.073 ≈ (3 + x)2 − (9 − x)

7.073 ≈ 9 + 6x + x2 − 9 + x 7.073 ≈ x2 + 7x

19.323 ≈ x2 + 7x + 49 —

4

19.323 ≈ ( x + 7 — 2 ) 2

± √—

19.323 ≈ x + 7 — 2

− 7 — 2 ± √

— 19.323 ≈ x

−3.5 ± √—

19.323 ≈ x

Reject the negative solution, −3.5 − √—

19.323 ≈ −7.896,

because the thickness must be positive. So, the thickness is

x ≈ −3.5 + √—

19.323 ≈ 0.896 centimeter.


74. 2x − 3 < 5

2x < 8

x < 4

1 3

4

5 7 x

75. 4 − 8y ≥ 12

−8y ≥ 8

y ≤ −1

−4 −2

−1

0 2 y

76. n —

3 + 6 > 1

n + 18 > 3

n > −15

−18 −16 −14 −12

−15

n

77. − 2s —

5 ≤ 8

−2s ≤ 40

s ≥ −20

−23 −21 −19 −17

−20

s

78. y

2 6

2

4

6

x

x = 4

g(x) = 6(x − 4)2(4, 0)

79. y

42−2

−4

−2

6

8

x

h(x) = 2x(x − 3)

(1.5, −4.5)

(0, 0) (3, 0)

x = 1.5

80. y

2−2−4

6

4

2

8

x

(−1, 4)

f(x) = x2 + 2x + 5

x = −1

81.

84−4−8

−200

−100

100

x

y

(−10, 0)

(1, −242)

(12, 0)

f(x) = 2(x + 10)(x − 12)

x = 1

3.1–3.3 What Did You Learn? (p. 119)

1. The quilt is originally 4 feet by 5 feet but, with the border

added on, the new dimensions are 4 + 2x and 5 + 2x. It also

states that the border will be uniform and exactly 10 square

feet. Multiplying the new length and width and then

subtracting the original area of 20 square feet equals the area

of the border, 10 square feet. Setting up this equation allows

you to solve for the width, x, of the added border.

2. Because there is no constant term in the equation in

Exercise 67, it is easier to fi nd the zeros. Simply factor,

set each factor to zero, and solve. In Exercise 63, the fi rst

step would be factoring a from the fi rst two terms before

completing the square. Once in vertex form, the equation can

then be solved. Exercise 63 requires more steps.


Chapter 3

3.1–3.3 Quiz (p. 120)

1. The solution is x = 5.

Check: x2 − 10x + 25 = 0

(5)2 − 10(5) + 25 =?

0

25 − 50 + 25 =?

0

0 = 0 ✓

2. The solutions are x = 2 and x = 4.

Check:

2x2 + 16 = 12x 2x2 + 16 = 12x

2(2)2 + 16 =?

12(2) 2(4)2 + 16 =?

12(4)

8 + 16 =?

24 32 + 16 =?

48

24 = 24 ✓ 48 = 48 ✓

3. The solutions are x = −4 and x = 2.

Check:

x2 = −2x + 8 x2 = −2x + 8

(−4)2 =?

−2(−4) + 8 22 =?

−2(2) + 8

16 =?

8 + 8 4 =?

−4 + 8

16 = 16 ✓ 4 = 4 ✓

4. Use the square root method because the equation can be

written in the form u2 = d.

2x2 − 15 = 0 2x2 = 15

x2 = 15 —

2

x = ± √—

30 —

2


30 —

2 and x = √

— 30 —

2 .

5. Use the factoring method because the left-hand side factors.

3x2 − x − 2 = 0 (3x + 2)(x − 1) = 0 3x + 2 = 0 or x − 1 = 0

x = − 2 — 3 or x = 1

The solutions are x = − 2 — 3 and x = 1.

6. Use the square root method because the equation is of the

form u2 = d.

(x + 3)2 = 8 x + 3 = ±2 √

— 2

x = −3 ± 2 √—

2

The solutions are x = −3 − 2 √—

2 and x = −3 + 2 √—

2 .



7x = 14 −6 = y x = 2 y = −6

So, x = 2 and y = −6.

8. (2 + 5i) + (−4 + 3i) = (2 − 4) + (5 + 3)i

= −2 + 8i

9. (3 + 9i) − (1 − 7i) = (3 − 1) + (9 + 7)i

= 2 + 16i

10. (2 + 4i)(−3 − 5i) = −6 − 10i − 12i − 20i2

= −6 − 22i − 20(−1)

= 14 − 22i

11. 0 = 9x2 + 2 −2 = 9x2

− 2 — 9 = x2

±i √

— 2 —

3 = x

The zeros of the function are x = −i √

— 2 —

3 and x = i √

— 2 —

3 . The

graph of the function does not intercept the x-axis because

the function has imaginary numbers as zeros.

12. x2 − 6x + 10 = 0 x2 − 6x = −10

x2 − 6x + 9 = −10 + 9 (x − 3)2 = −1

x − 3 = ±i

x = 3 ± i The solutions are x = 3 − i and x = 3 + i.

13. x2 + 12x + 4 = 0 x2 + 12x = −4

x2 + 12x + 36 = −4 + 36

(x + 6)2 = 32

x + 6 = ±4 √—

2

x = −6 ± 4 √—

2

The solutions are x = −6 − 4 √—

2 and x = −6 + 4 √—

2 .

14. 4x(x + 6) = −40

x(x + 6) = −10

x2 + 6x = −10

x2 + 6x + 9 = −10 + 9 (x + 3)2 = −1

x + 3 = ±i

x = −3 ± i The solutions are x = −3 − i and x = −3 + i.


Chapter 3

15. y = x2 − 10x + 4 y + ? = (x2 − 10x + ?) + 4 y + 25 = (x2 − 10x + 25) + 4 y + 25 = (x − 5)2 + 4 y = (x − 5)2 − 21

The vertex form of the function is y = (x − 5)2 − 21. The

vertex is (5, −21).

16. a. The area of the existing patio is 20 ⋅ 30 = 600 square feet.

b. An equation that will model the area of the new patio is

600 + 464 = (20 + x)(30 + x).

c. Solve the equation in part (b).

600 + 464 = (20 + x)(30 + x)

1064 = 600 + 50x + x2

464 = x2 + 50x

464 + 625 = x2 + 50x + 625

1089 = (x + 25)2

±33 = x + 25

−25 ± 33 = x Reject the negative solution, −25 − 33 = −58, because

lengths are positive. So, the patio should be extended −25 + 33 = 8 feet.





Impedance of circuit = 7 + 5i + (−2i) = 7 + 3i

The impedance of the circuit is (7 + 3i) ohms.

18. a. Write the function in vertex form.

h = −16t2 + 32t + 4 h = −16(t2 − 2t) + 4 h + ? = −16(t2 − 2t + ?) + 4 h + (−16)(1) = −16(t2 − 2t + 1) + 4 h − 16 = −16(t − 1)2 + 4 h = −16(t − 1)2 + 20

The vertex is (1, 20). So, the maximum height of the

birdie is 20 feet.

b. Find the zeros of the function.

0 = −16(t − 1)2 + 20

−20 = −16(t − 1)2

1.25 = (t − 1)2

± √—

1.25 = t − 1 1 ± √

— 1.25 = t

Reject the negative solution, 1 − √—

1.25 ≈ − 0.12

because time must be positive. So, the birdie is in the air

for 1 + √—

1.25 ≈ 2.12 seconds.


1.

ax2 + bx + c = 0 Write the equation.

ax2 + bx = −c Subtract c from each side.

x2 + b —

a x = − c —

a Divide each side by a.

x2 + b —

a x + ( b —

2a )

2

= − c — a + ( b —

2a )

2

Add ( b — 2a

) 2

to each side.

x2 + b —

a x + ( b —

2a )

2

= − 4ac —

4a2 + b2

— 4a2

Find common

denominator.

( x + b —

2a )

2

= b2 − 4ac

— 4a2

Factor and add fractions.

x + b —

2a = ± √—

b2 − 4ac —

4a2

Take square root of

each side.

x = − b — 2a

± √—

b2 − 4ac —

2 ∣ a ∣ Subtract b —

2a from

each side.

x = −b ± √

— b2 − 4ac ——

2a Add fractions.

2. a. x2 − 4x + 3 = 0

x = −b ± √

— b2 − 4ac ——

2a

x = −(−4) ± √

—— (−4)2 − 4(1)(3) ———

2(1)

x = 4 ± √

— 4 —

2

x = 4 ± 2

— 2

x = 2 ± 1

So, the solutions are x = 1 and x = 3.

b. x2 − 2x + 2 = 0

x = −b ± √

— b2 − 4ac ——

2a

x = −(−2) ± √

—— (−2)2 − 4(1)(2) ———

2(1)

x = 2 ± √

— −4 —

2

x = 2 ± 2i

— 2

x = 1 ± i

So, the solutions are x = 1 − i and x = 1 + i.


Chapter 3

c. x2 + 2x − 3 = 0

x = −b ± √

— b2 − 4ac ——

2a

x = −2 ± √——

22 − 4(1)(−3) ——

2(1)

x = −2 ± √

— 16 —

2

x = −2 ± 4

— 2

x = −1 ± 2

So, the solutions are x = 1 and x = −3.

d. x2 + 4x + 4 = 0

x = −b ± √

— b2 − 4ac ——

2a

x = −4 ± √

—— 42 − 4(1)(4) ——

2(1)

x = −4 ± √

— 0 —

2

x = − 4 —

2

x = −2

So, the solution is x = −2.

e. x2 − 6x + 10 = 0

x = −b ± √

— b2 − 4ac ——

2a

x = −(−6) ± √

—— (−6)2 − 4(1)(10) ———

2(1)

x = 6 ± √

— −4 —

2

x = 6 ± 2i

— 2

x = 3 ± i

So, the solutions are x = 3 − i and x = 3 + i.

f. x2 + 4x + 6 = 0

x = −b ± √

— b2 − 4ac ——

2a

x = −4 ± √

—— 42 − 4(1)(6) ——

2(1)

x = −4 ± √

— −8 —

2

x = −4 ± 2i √

— 2 —

2

x = −2 ± i √—

2

So, the solutions are x = −2 − i √—

2 and x = −2 + i √—

2 .

3. You derive the quadratic formula by using the process

of completing the square to solve the general form of a

quadratic equation.

4. Graphing: Use the graph to fi nd the x-intercepts; these are

the solutions to the equation.

Square root: If there is a square with no x-terms, then you

can take the square root of each side of the equation to solve.

Factoring: Factor the quadratic part of the equation and set

each factor equal to zero to solve for the variable.

Completing the square: Add a new term to each side of the

equation so that the quadratic part of the equation factors

into a perfect square. Then use square roots.

Quadratic Formula: Use the coeffi cients from the quadratic

equation in the formula to directly compute the solutions.


1. x2 − 6x + 4 = 0

x = −b ± √

— b2 − 4ac ——

2a

x = −(−6) ± √

—— (−6)2 − 4(1)(4) ———

2(1)

x = 6 ± √

— 20 —

2

x = 6 ± 2 √

— 5 —

2

x = 3 ± √—

5

So, the solutions are x = 3 − √—

5 and x = 3 + √—

5 .

2. 2x2 + 4 = −7x

2x2 + 7x + 4 = 0

x = −b ± √

— b2 − 4ac ——

2a

x = −7 ± √

—— 72 − 4(2)(4) ——

2(2)

x = −7 ± √

— 17 —

4

So, the solutions are x = −7 − √

— 17 —

4 and x =

−7 + √—

17 —

4 .

3. 5x2 = x + 8

5x2 − x − 8 = 0

x = −b ± √

— b2 − 4ac ——

2a

x = −(−1) ± √

—— (−1)2 − 4(5)(−8) ———

2(5)

x = 1 ± √

— 161 —

10

So, the solutions are x = 1 − √

— 161 —

10 and x =

1 + √—

161 —

10 .


Chapter 3

4. x2 + 41 = −8x

x2 + 8x + 41 = 0

x = −b ± √

— b2 − 4ac ——

2a

x = −8 ± √

—— 82 − 4(1)(41) ——

2(1)

x = −8 ± √

— −100 ——

2

x = −8 ± 10i

— 2

x = −4 ± 5i

So, the solutions are x = −4 − 5i and x = −4 + 5i .

5. − 9x2 = 30x + 25

−9x2 − 30x − 25 = 0

x = −b ± √

— b2 − 4ac ——

2a

x = −(−30) ± √

—— (−30)2 − 4(−9)(−25) ———

2(−9)

x = 30 ± √

— 0 —

−18

x = 30

— −18

x = − 5 — 3

So, the solution is x = − 5 — 3 .

6. 5x − 7x2 = 3x + 4

−7x2 + 2x − 4 = 0

x = −b ± √

— b2 − 4ac ——

2a

x = −2 ± √

—— 22 − 4(−7)(−4) ———

2(−7)

x = −2 ± √

— −108 ——

−14

x = −2 ± 6i √

— 3 —

−14

x = 1 ± 3i √

— 3 —

7

So, the solutions are x = 1 − 3i √

— 3 —

7 and x =

1 + 3i √—

3 —

7 .

7. Equation: 4x2 + 8x + 4 = 0

Discriminant: b2 − 4ac = 82 − 4(4)(4) = 0

The equation has one real solution.

Solution: x = −b ± √

— b2 − 4ac ——

2a

= −8 ± √

— (0) —

2(4)

= −1

8. Equation: 1 —

2 x2 + x − 1 = 0

Discriminant: b2 − 4ac = 12 − 4 ( 1 — 2 ) (−1) = 3

The equation has two real solutions.

Solutions: x = −b ± √

— b2 − 4ac ——

2a

= −1 ± √

— 3 —

2 ( 1 — 2 )

= −1 ± √—

3

9. Equation: 5x2 − 8x + 13 = 0

Discriminant: b2 − 4ac = (−8)2 − 4(5)(13) = −196

The equation has two imaginary solutions.


— b2 − 4ac ——

2a

= −(−8) ± √

— −196 ——

2(5)

= 8 ± 14i

— 10

= 4 ± 7i

— 5

10. Equation: 7x2 − 3x − 6 = 0

Discriminant: b2 − 4ac = (−3)2 − 4(7)(−6) = 177



— b2 − 4ac ——

2a

= −(−3) ± √

— 177 ——

2(7)

= 3 ± √

— 177 —

14

11. Equation: 4x2 + 6x + 9 = 0

Discriminant: b2 − 4ac = 62 − 4(4)(9) = −108



— b2 − 4ac ——

2a

= −6 ± √

— −108 ——

2(4)

= −6 ± 6i √

— 3 —

8

= −3 ± 3i √

— 3 —

4


Chapter 3

12. Equation: −5x2 + 10x − 5 = 0

Discriminant: b2 − 4ac = 102 − 4(−5)(−5) = 0



— b2 − 4ac ——

2a =

−10 ± √—

0 —

2(−5) = 1

13. Sample answer: For a quadratic equation to have two real

solutions, the discriminant must be positive. So, set the

discriminant equal to any positive number.

b2 − 4ac = 13

32 − 4ac = 13

9 − 4ac = 13

− 4ac = 4

ac = −1

Because ac = −1, choose two integers whose product is −1,

such as a = 1 and c = −1. So, one possible equation is

x2 + 3x − 1 = 0.

14. Because the ball is thrown, use the model

h = −16t2 + v 0t + h0. To fi nd how long the ball is in the air,

solve for t when h = 3.

h = −16t2 + v0t + h0

3 = −16t2 + 40t + 4

0 = −16t2 + 40t + 1

This equation is not factorable, and completing the square

would result in fractions. So, use the Quadratic Formula to

solve the equation.

t = −40 ± √

—— 402 − 4(−16)(1) ———

2(−16)

t = −40 ± √

— 1664 ——

−32

t ≈ −0.025 or t ≈ 2.52

Reject the negative solution, −0.025, because time cannot be

negative. So, the ball is in the air for about 2.52 seconds.



1. When a, b, and c are real numbers such that a ≠ 0, the

solutions of the quadratic equation

ax2 + bx + c = 0 are x = −b ± √

— b2 − 4ac ——

2a .

2. You can use the discriminant of a quadratic equation to

determine the number and type of solutions of the equation.

3. If the discriminant of a quadratic equation is negative, then

the equation has two imaginary solutions.

4. You can use the process of completing the square and

the Quadratic Formula to solve any quadratic equation.

Complete the square when a = 1 and b is an even number.

Use the Quadratic Formula when a ≠ 1, or b is an odd number.


5. x2 − 4x + 3 = 0

x = −b ± √

— b2 − 4ac ——

2a

x = −(−4) ± √

—— (−4)2 − 4(1)(3) ———

2(1)

x = 4 ± √

— 4 —

2

x = 4 ± 2

— 2

x = 2 ± 1

So, the solutions are x = 1 and x = 3.

6

−6

−2

10

ZeroX=1 Y=0

6. 3x2 + 6x + 3 = 0

x = −b ± √

— b2 − 4ac ——

2a

x = −6 ± √

—— 62 − 4(3)(3) ——

2(3)

x = −6 ± 0

— 6

x = −1


4

−4

−6

10

ZeroX=-1 Y=0

7. x2 + 6x + 15 = 0

x = −b ± √

— b2 − 4ac ——

2a

x = −6 ± √

—— 62 − 4(1)(15) ——

2(1)

x = −6 ± √

— −24 ——

2

x = −6 ± 2i √

— 6 —

2

x = −3 ± i √—

6

So, the solutions are x = −3 − i √—

6 and x = −3 + i √—

6 .

40

−4

20


Chapter 3

8. 6x2 − 2x + 1 = 0

x = −b ± √

— b2 − 4ac ——

2a

x = −(−2) ± √

—— (−2)2 − 4(6)(1) ———

2(6)

x = 2 ± √

— −20 —

12

x = 2 ± 2i √

— 5 —

12

x = 1 ± i √

— 5 —

6

So, the solutions are x = 1 + i √

— 5 —

6 and x =

1 − i √—

5 —

6 .

4

−2

−4

8

9. x2 − 14x = −49

x2 − 14x + 49 = 0

x = −b ± √

— b2 − 4ac ——

2a

x = −(−14) ± √

—— (−14)2 − 4(1)(49) ———

2(1)

x = 14 ± 0

— 2

x = 7

So, the solution is x = 7.

11

−20

−3

40

ZeroX=7 Y=0

10. 2x2 + 4x = 30

2x2 + 4x − 30 = 0

x = −b ± √

— b2 − 4ac ——

2a

x = −4 ± √

—— 42 − 4(2)(−30) ——

2(2)

x = −4 ± √

— 256 —

4

x = −4 ± 16

— 4

So, the solutions are x = −5 10

−50

−10

10

ZeroX=-5 Y=0

and x = 3.

11. 3x2 + 5 = −2x

3x2 + 2x + 5 = 0

x = −b ± √

— b2 − 4ac ——

2a

x = −2 ± √

—— 22 − 4(3)(5) ——

2(3)

x = −2 ± √

— −56 ——

6

x = −2 ± 2i √

— 14 ——

6

x = −1 ± i √

— 14 —

3

So, the solutions are x = −1 − i √

— 14 —

3 and x =

−1 + i √—

14 —

3 .

40

−4

10

12. −3x = 2x2 − 4

2x2 + 3x − 4 = 0

x = −b ± √

— b2 − 4ac ——

2a

x = −3 ± √

—— 32 − 4(2)(−4) ——

2(2)

x = −3 ± √

— 41 —

4

So, the solutions are x = −3 − √

— 41 —

4 and x =

−3 + √—

41 —

4 .

10

−10

−10

10

ZeroX=.8507810 Y=0

13. −10x = −25 − x2

x2 − 10x + 25 = 0

x = −b ± √

— b2 − 4ac ——

2a

x = −(−10) ± √

—— (−10)2 − 4(1)(25) ———

2(1)

x = 10 ± √

— 0 —

2

x = 10 ± 0

— 2

x = 5

So, the solution is x = 5.

8

−6

0

20

Y=0ZeroX=5


Chapter 3

14. −5x2 − 6 = −4x

−5x2 + 4x − 6 = 0

x = −b ± √

— b2 − 4ac ——

2a

x = −4 ± √

—— 42 − 4(−5)(−6) ———

2(−5)

x = −4 ± √

— −104 ——

−10

x = −4 ± 2i √

— 26 ——

−10

x = 2 ± i √

— 26 —

5

So, the solutions are x = 2 − i √

— 26 —

5 and x =

2 + i √—

26 —

5 .

4

−30

−40

15. −4x2 + 3x = −5

−4x2 + 3x + 5 = 0

x = −b ± √

— b2 − 4ac ——

2a

x = −3 ± √

—— 32 − 4(−4)(5) ——

2(−4)

x = −3 ± √

— 89 —

−8

So, the solutions are x = 3 − √

— 89 —

8 and x =

3 + √—

89 —

8 .

4

−10

−4

6

ZeroY=0X=-.804247

16. x2 + 121 = −22x

x2 + 22x + 121 = 0

x = −b ± √

— b2 − 4ac ——

2a

x = −22 ± √

—— 222 − 4(1)(121) ———

2(1)

x = −22 ± √

— 0 —

2

x = −11


2

−4

−20

8

ZeroY=0X=-11

17. −z2 = −12z + 6

−z2 + 12z − 6 = 0

z = −b ± √

— b2 − 4ac ——

2a

z = −12 ± √

—— 122 − 4(−1)(−6) ———

2(−1)

z = −12 ± √

— 120 ——

−2

z = −12 ± 2 √

— 30 ——

−2

z = 6 ± √—

30

So, the solutions are z = 6 − √—

30 and z = 6 + √—

30 .

18

−20

−6

40

ZeroY=0X=11.477225

18. −7w + 6 = −4w2

4w2 − 7w + 6 = 0

w = −b ± √

— b2 − 4ac ——

2a

w = −(−7) ± √

—— (−7)2 − 4(4)(6) ———

2(4)

w = 7 ± √

— −47 —

8

w = 7 ± i √

— 47 —

8

So, the solutions are w = 7 − i √

— 47 —

8 and w =

7 + i √—

47 —

8 .

4

−6

−4

20

19. Equation: x2 + 12x + 36 = 0




— b2 − 4ac ——

2a =

−12 ± √—

0 —

2(1) = −6


Chapter 3

20. Equation: x2 − x + 6 = 0




— b2 − 4ac ——

2a

x = −(−1) ± √

— −23 ——

2(1)

x = 1 ± i √

— 23 —

2

21. Equation: 4n2 − 4n − 24 = 0

Discriminant: b2 − 4ac = (−4)2 − 4(4)(−24) = 400


Solutions: n = −b ± √

— b2 − 4ac ——

2a

n = −(−4) ± √

— 400 ——

2(4)

n = 4 ± 20

— 8

n = −2 and n = 3

22. Equation: −x2 + 2x + 12 = 0

Discriminant: b2 − 4ac = 22 − 4(−1)(12) = 52



— b2 − 4ac ——

2a

x = −2 ± √

— 52 —

2(1)

x = −1 ± √—

13

23. Equation: 4x2 − 5x + 10 = 0




— b2 − 4ac ——

2a

x = −(−5) ± √

— −135 ——

2(4)

x = 5 ± 3i √

— 15 —

8

24. Equation: p2 + 18p + 81 = 0



Solution: p = −b ± √

— b2 − 4ac ——

2a =

−18 ± √—

0 —

2(1) = −9

25. Equation: 3x2 + 24x + 48 = 0




— b2 − 4ac ——

2a =

−24 ± √—

0 —

2(3) = −4

26. Equation: −2x2 − x − 6 = 0

Discriminant: b2 − 4ac = (−1)2 − 4(−2)(−6) = −47



— b2 − 4ac ——

2a

x = −(−1) ± √

— −47 ——

2(−2)

x = −1 ± i √

— 47 —

4

27. A; 2x2 − 16x + 50 = 0

x = −b ± √

— b2 − 4ac ——

2a

x = −(−16) ± √

—— (−16)2 − 4(2)(50) ———

2(2)

x = 16 ± √

— −144 ——

4

x = 16 ± 12i

— 4

x = 4 ± 3i

28. A; The discriminant is 72 − 4(1)(11) = 5, so there are two

real solutions.

29. C; The discriminant is (−6)2 − 4(1)(25) = −64, so the

graph has no x-intercepts.

30. D; The discriminant is (−20)2 − 4(2)(50) = 0, so the graph

has one x-intercept.

31. A; The discriminant is 62 − 4(3)(−9) = 144, so the graph

has two x-intercepts and a y-intercept of −9.

32. B; The discriminant is (−10)2 − 4(5)(−35) = 800, so the

graph has two x-intercepts and a y-intercept of −35.

33. The square root is an imaginary number, not a real number.

x2 + 10x + 74 = 0

x = −10 ± √

—— 1002 − 4(1)(74) ———

2(1)

= −10 ± √

— −196 ——

2

= −10 ± 14i

— 2

= −5 ± 7i

= −5 − 7i or −5 + 7i


Chapter 3

34. The equation was not written in standard form when the

quadratic equation was used.

x2 + 6x + 8 = 2

x2 + 6x + 6 = 0

x = −6 ± √

—— 62 − 4(1)(6) ——

2(1)

= −6 ± √

— 12 —

2

= −6 ± 2 √

— 3 —

2

= −3 ± √—

3

= −3 − √—

3 or −3 + √—

3

35. Sample answer: For a quadratic equation to have two

imaginary solutions, the discriminant must be negative. So,

set the discriminant equal to any negative number.

b2 − 4ac = −12

42 − 4ac = −12

16 − 4ac = −28

−4ac = −28

ac = 7

Because ac = 7, choose two integers whose product is 7,

such as a = 1 and c = 7. So, one possible equation is

x2 + 4x + 7 = 0.




b2 − 4ac = 24

62 − 4ac = 24

36 − 4ac = 24

−4ac = −12

ac = 3


such as a = −3 and c = −1. So, one possible equation is

−3x2 + 6x − 1 = 0.




b2 − 4ac = 24

(−8)2 − 4ac = 24

64 − 4ac = 24

−4ac = −40

ac = 10

Because ac = 10, choose two integers whose product is

10, such as a = 2 and c = 5. So, one possible equation is

2x2 − 8x + 5 = 0.

38. Sample answer: For a quadratic equation to have one real

solution, the discriminant must equal zero. So, set the

discriminant equal to 0.

b2 − 4ac = 0

(−6)2 − 4ac = 0

36 − 4ac = 0

−4ac = −36

ac = 9


such as a = −3 and c = −3. So, one possible equation is

−3x2 − 6x − 3 = 0.

39. Sample answer: Rewrite the equation as ax2 + 10x − c = 0.

For a quadratic equation to have one real solution, the

discriminant must equal zero. So, set the discriminant

equal to 0.

b2 − 4ac = 0

102 − 4a(−c) = 0

100 + 4ac = 0

4ac = −100

ac = −25

Because ac = −25, choose two integers whose product is

−25, such as a = 1 and c = −25. So, one possible equation

is x2 + 10x + 25 = 0.

40. Sample answer: Rewrite the equation as ax2 − 4x + c = 0.

For a quadratic equation to have two imaginary solutions, the

discriminant must be negative. So, set the discriminant equal

to any negative number.

b2 − 4ac = −16

(−4)2 − 4ac = −16

16 − 4ac = −32

−4ac = −32

ac = 8


such as a = 1 and c = 8. So, one possible equation is

x2 − 4x + 8 = 0.

41. x = −8 ± √

— −176 ——

−10

−10x = −8 ± √—

−176

−10x + 8 = ± √—

−176

(−10x + 8)2 = −176

100x2 − 160x + 64 = −176

100x2 − 160x + 240 = 0

−5x2 + 8x − 12 = 0

So, an equation is −5x2 + 8x − 12 = 0.


Chapter 3

42. x = 15 ± √

— −215 ——

22

22x = 15 ± √—

−215

22x − 15 = ± √—

−215

(22x − 15)2 = −215

484x2 − 660x + 225 = −215

484x2 − 660x + 440 = 0

11x2 − 15x + 10 = 0

So, an equation is 11x2 − 15x + 10 = 0.

43. x = −4 ± √

— −124 ——

−14

−14x = −4 ± √—

−124

−14x + 4 = ± √—

−124

(−14x + 4)2 = −124

196x2 − 112x + 16 = −124

196x2 − 112x + 140 = 0

−7x2 + 4x − 5 = 0

So, an equation is −7x2 + 4x − 5 = 0.

44. x = −9 ± √

— 137 —

4

4x = −9 ± √—

137

4x + 9 = ± √—

137

(4x + 9)2 = 137

16x2 + 72x + 81 = 137

16x2 + 72x − 56 = 0

2x2 + 9x − 7 = 0

So, an equation is 2x2 + 9x − 7 = 0.

45. x = −4 ± 2

— 6

6x = −4 ± 2

6x + 4 = ±2

(6x + 4)2 = 4

36x2 + 48x + 16 = 4

36x2 + 48x + 12 = 0

3x2 + 4x + 1 = 0

So, an equation is 3x2 + 4x + 1 = 0.

46. x = 2 ± 4

— −2

−2x = 2 ± 4

−2x − 2 = ±4

(−2x − 2)2 = 16

4x2 + 8x + 4 = 16

4x2 + 8x − 12 = 0

−x2 − 2x + 3 = 0

So, an equation is −x2 − 2x + 3 = 0.

47. Quadratic Formula:

3x2 − 21 = 3 3x2 − 24 = 0

x = 0 ± √——

02 − 4(3)(−24) ——

2(3)

x = ± √—

288 —

6

x = ±12 √—

2 —

6

x = ±2 √—

2

Square roots:

3x2 − 21 = 3 3x2 = 24

x2 = 8 x = ±2 √

— 2

Sample answer: Square roots is preferred because the

equation can be written in the form u2 = d.


5x2 + 38 = 3 5x2 + 35 = 0

x = 0 ± √——

02 − 4(5)(35) ——

2(5)

x = ± √—

−700 —

10

x = ±10i √—

7 —

10

x = ± i √—

7

Square roots:

5x2 + 38 = 3 5x2 = −35

x2 = −7

x = ±i √—

7

Sample answer: Square roots is preferred because the

equation can be written in the form u2 = d.


Chapter 3


2x2 − 54 = 12x

2x2 − 12x − 54 = 0

x = −(−12) ± √——

(−12)2 − 4(2)(−54) ———

2(2)

x = 12 ± √—

576 —

4

x = 12 ± 24 —

4

x = 3 ± 6 x = 9 or x = −3

Factoring:

2x2 − 54 = 12x

2x2 − 12x − 54 = 0 x2 − 6x − 27 = 0 (x − 9)(x + 3) = 0 x − 9 = 0 or x + 3 = 0 x = 9 or x = −3

Sample answer: Factoring is preferred because the equation

can be factored.


x2 = 3x + 15

x2 − 3x − 15 = 0

x = −(−3) ± √——

(−3)2 − 4(1)(−15) ———

2(1)

x = 3 ± √—

69 —

2

Completing the square:

x2 = 3x + 15

x2 − 3x + 9 — 4 = 15 + 9 —

4

( x − 3 — 2

) 2

= 69 —

4

x − 3 — 2 = ± √

— 69 —

2

x = 3 ± √—

69 —

2

Sample answer: The Quadratic Formula is preferred because

b is not an even number, the equation cannot be factored, and

it cannot be easily written in the form u2 = d.


x2 − 7x + 12 = 0

x = −(−7) ± √——

(−7)2 − 4(1)(12) ———

2(1)

x = 7 ± √—

1 —

2

x = 7 ± 1 —

2

x = 4 or x = 3 Factoring:

x2 − 7x + 12 = 0 (x − 3)(x − 4) = 0 x − 3 = 0 or x − 4 = 0 x = 3 or x = 4 Sample answer: Factoring is preferred because the equation

can be factored.


x2 + 8x − 13 = 0

x = −8 ± √——

82 − 4(1)(−13) ——

2(1)

x = −8 ± √—

116 —

2

x = −8 ± 2 √—

29 —

2

x = −4 ± √—

29


x2 + 8x − 13 = 0 x2 + 8x = 13

x2 + 8x + 16 = 13 + 16

(x + 4)2 = 29

x + 4 = ± √—

29

x = −4 ± √—

29

Sample answer: Completing the square is preferred because

a = 1 and b is an even number.


Chapter 3


5x2 − 50x = −135

5x2 − 50x + 135 = 0

x = −(−50) ± √——

(−50)2 − 4(5)(135) ———

2(5)

x = 50 ± √—

−200 ——

10

x = 50 ± 10i √—

2 ——

10

x = 5 ± i √—

2


5x2 − 50x = −135

x2 − 10x = −27

x2 − 10x + 25 = −27 + 25

(x − 5)2 = −2

x − 5 = ± i √—

2

x = 5 ± i √—

2

Sample answer: Completing the square is preferred because

you fi rst factor out a 5, and then a = 1 and b is an even number.


8x2 + 4x + 5 = 0

x = −4 ± √——

42 − 4(8)(5) ——

2(8)

x = −4 ± √—

−144 ——

16

x = −4 ± 12i —

16

x = −1 ± 3i —

4


8x2 + 4x + 5 = 0 8x2 + 4x = −5

x2 + 1 — 2 x = − 5 —

8

x2 + 1 — 2 x + 1 —

16 = − 5 —

8 + 1 —

16

( x + 1 — 4 )

2

= − 9 — 16

x + 1 — 4 = ± 3 —

4 i

x = − 1 — 4

± 3 — 4 i

The Quadratic Formula is preferred because a ≠ 1, the

equation cannot be factored, and it cannot be easily written

in the form u2 = d.


−3 = 4x2 + 9x

4x2 + 9x + 3 = 0

x = −9 ± √——

92 − 4(4)(3) ——

2(4)

x = 9 ± √—

33 —

8


4x2 + 9x = −3

4 ( x2 + 9 — 4 x ) = −3

4 ( x2 + 9 — 4 x + 81

— 64

) = −3 + 81 —

16

4 ( x + 9 — 8 )

2

= 33 —

16

( x + 9 — 8 )

2

= 33 —

64

x + 9 — 8 = ± √

— 33 —

8

x = − 9 — 8 ± √

— 33 —

8

The Quadratic Formula is preferred because a ≠ 1, b is

not an even number, the equation cannot be factored, and it

cannot be easily written in the form u2 = d.


−31x + 56 = −x2

x2 + 31x + 56 = 0

x = −31 ± √——

312 − 4(1)(56) ——

2(1)

x = −31 ± √—

737 ——

2


−31x + 56 = −x2

x2 + 31x = −56

x2 + 31x + 961 —

4 = −56 + 961

— 4

( x + 31 —

2 )

2

= 737 —

4

x + 31 —

2 = ± √

— 737 —

2

x = −31 ± √—

737 ——

2

The Quadratic Formula is preferred because b is not an even

number, the equation cannot be factored, and it cannot be

easily written in the form u2 = d.


Chapter 3


x2 = 1 − x x2 + x − 1 = 0

x = −1 ± √——

12 − 4(1)(−1) ——

2(1)

x = −1 ± √—

5 —

2


x2 = 1 − x x2 + x = 1

x2 + x + 1 — 4

= 1 + 1 — 4

( x + 1 — 2 )

2

= 5 — 4

x + 1 — 2

= ± √—

5 —

2

x = −1 ± √—

5 —

2

The Quadratic Formula is preferred because b is not an even

number, the equation cannot be factored, and it cannot be

easily written in the form u2 = d.


9x2 + 36x + 72 = 0

x = −36 ± √——

362 − 4(9)(72) ——

2(9)

x = −36 ± √—

−1296 ——

18

x = −36 ± 36i —

18

x = −2 ± 2i


9x2 + 36x + 72 = 0 x2 + 4x = −8

x2 + 4x + 4 = −8 + 4 (x + 2)2 = −4

x + 2 = ±2i

x = −2 ± 2i

Completing the square is preferred. Factor out 9, and a = 1

and b is an even number.

59. The equation that models the area of the rectangle is

24 = (2x − 9)(x + 2). Solve the equation.

24 = (2x − 9)(x + 2)

24 = 2x2 − 5x − 18

0 = 2x2 − 5x − 42

x = −(−5) ± √——

(−5)2 − 4(2)(−42) ———

2(2)

x = 5 ± √—

361 —

4

x = 5 ± 19 —

4

x = 6 or − 7 — 2

Reject the negative solution, − 7 — 2 , because the length must be

positive. So, x = 6.

60. The equation that models the area of the triangle is

8 = 1 — 2 (3x − 7)(x + 1). Solve the equation.

8 = 1 — 2 (3x − 7)(x + 1)

16 = 3x2 − 4x − 7 0 = 3x2 − 4x − 23

x = −(−4) ± √——

(−4)2 − 4(3)(−23) ———

2(3)

x = 4 ± √—

292 —

6

x = 4 ± 2 √—

73 —

6

x = 4 − 2 √—

73 —

6 or x =

4 + 2 √—

73 —

6

Reject the negative solution, 4 − 2 √

— 73 —

6 ≈ −2.18, because

the length must be positive. So, x = 4 + 2 √—

73 —

6 ≈ 3.51.

61. Because the ball is thrown, use the model

h = −16t2 + v 0t + h 0. To fi nd how long the ball is in the air,


h = −16t2 + v0t + h0

3 = −16t2 + 90t + 7 0 = −16t2 + 90t + 4 This equation is not factorable, and completing the square


solve the equation.

t = −90 ± √——

902 − 4(−16)(4) ———

2(−16)

t = −90 ± √—

8356 —— −32

t ≈ −0.044 or t ≈ 5.670

Reject the negative solution, −0.044, because the ball’s time

in the air cannot be negative. So, the ball is in the air for

about 5.67 seconds.


Chapter 3

62. It is not possible for a and c to be integers, but a and c can

be rational numbers. For ax2 + 5x + c = 0 to have one real

solution, the discriminant must be 0.

b2 − 4ac = 0 25 − 4ac = 0 −4ac = −25

ac = 25 —

4

c = 25 —

4a

Because 4 is not a factor of 25, then a or c will be a rational

number.

63. Because the ball is hit down, use the model

h = −16t2 + v0t + h0. To fi nd how long the ball is in the air,


h = −16t2 + v0t + h0

0 = −16t2 − 55t + 10



solve the equation.

t = −(−55) ± √——

(−55)2 − 4(−16)(10) ———

2(−16)

t = 55 ± √—

3665 —— −32

t ≈ −3.6 or t ≈ 0.17

Reject the negative solution, −3.6, because the ball’s time in

the air cannot be negative. So, the ball is in the air for about

0.17 second.

64. a. Because the arrow is shot parallel to the ground, use the

model h = −16t2 + h0. To fi nd how long the arrow is in

the air, solve for t when h = 3.

h = −16t2 + h0

3 = −16t2 + 5 0 = −16t2 + 2 16t2 = 2 t2 = 1 —

8

t = ± √

— 2 —

4

t ≈ −0.35 or t ≈ 0.35

Reject the negative solution, −0.35, because the arrow’s

time in the air cannot be negative. So, the arrow is in the

air for about 0.35 second.

b. Square roots was used because the equation can be written

in the form u2 = d.

65. a.

70

203

0

y = −16x2 + 105x + 30

y = −16x2 + 100x + 30

Both rockets start from the same height, but your friend’s

rocket does not go as high and lands about a half of a

second earlier.

b. The equation that models your rocket is

h = −16t2 + 105t + 30.

−16t2 + 105t + 30 = 119

−16t2 + 105t − 89 = 0 16t2 − 105t + 89 = 0 (t − 1)(16t − 89) = 0 t − 1 = 1 or 16t − 89 = 0

t = 1 or t = 89 —

16 = 5.5625

Your rocket is 119 feet above the ground after 1 second

and after 5.5625 seconds. These answers are reasonable

because 1 + 5.6

— 2 = 3.3, which is the axis of symmetry.

66. a. To fi nd which year there were 65 million tablet computers

sold, solve for t when A = 65.

65 = 4.5t2 + 43.5t + 17

0 = 4.5t2 + 43.5t − 48



solve the equation.

t = −43.5 ± √——

43.52 − 4(4.5)(−48) ———

2(4.5)

t = −43.5 ± √—

2756.25 ——

9

t = −10. — 6 or t = 1

Reject the negative solution, −10. — 6 , because the time in

years cannot be negative. So, the year 65 million tablet

computers were sold was 2010 + 1 = 2011.

b. The average rate of change from 2010 to 2012 is

122 − 17

— 2 − 0

= 52.5.

This means that from the year 2010 to 2012 the number

of tablet computers sold increased about 52.5 million each

year.

c. The model will not be accurate with the development of a

new, innovative computer because sales will most likely

decline sharply once the new computer is for sale.


Chapter 3

67. a. Because the gannet is fl ying down, use the model

h = −16t2 + v0t + h0. To fi nd how long the fi sh has to

swim away, solve for t when h = 0.

h = −16t2 + v0t + h0

0 = −16t2 − 88t + 100



solve the equation.

t = 88 ± √——

(−88)2 − 4(−16)(100) ———

2(−16)

t = 88 ± √—

14,144 —— −32

t ≈ −6.5 or t ≈ 0.97

Reject the negative solution, −6.5, because the time the

fi sh has to swim away cannot be negative. So, the fi sh has

about 0.97 second to swim away.

b. Because the second gannet is fl ying down, use the model

h = −16t2 + v0t + h0. To fi nd how long the fi sh has to

swim away, solve for t when h = 0.

h = −16t2 + v0t + h0

0 = −16t2 − 70t + 84



solve the equation.

t = 70 ± √——

(−70)2 − 4(−16)(84) ———

2(−16)

t = 70 ± √—

10,276 —— −32

t ≈ −5.4 or t ≈ 0.98

Reject the negative solution, −5.4, because the time the

fi sh has to swim away cannot be negative. So, the fi sh has

about 0.98 second to swim away from the second gannet.

So, the fi rst gannet will reach the fi sh fi rst.

68. Your solution is correct. Although it is diffi cult to see in a

standard viewing window, the graph touches the x-axis in

two places.

69. The equation that represents the total area of the pool and

deck is 400 = (2x + 18)(2x + 9). Solve the equation for x.

400 = (2x + 18)(2x + 9)

400 = 4x2 + 54x + 162

0 = 4x2 + 54x − 238

0 = 2(2x2 + 27x − 119)

0 = 2(2x − 7)(x + 17)

2x − 7 = 0 or x + 17 = 0 2x = 7 or x = −17

x = 3.5

Reject the negative solution, −17, because the width of the

deck cannot be negative. So, the width of the deck should be

3.5 feet.

70. a. The discriminant is negative because the graph has no

x-intercept. So, the equation has two imaginary solutions.

b. The discriminant is positive because the graph has two

x-intercepts. So, the equation has two real solutions.

c. The discriminant is zero because the graph has one

x-intercept. So, the equation has one real solution.

71. a. ∣ x2 − 3x − 14 ∣ = 4 x2 − 3x − 14 = 4 or x2 − 3x − 14 = −4

x2 − 3x − 18 = 0 or x2 − 3x − 10 = 0 (x − 6)(x + 3) = 0 or (x − 5)(x + 2) = 0

x − 6 = 0 or x + 3 = 0 or x − 5 = 0 or x + 2 = 0 x = 6 or x = −3 or x = 5 or x = −2

Check:

x = 6: ∣ 62 − 3(6) − 14 ∣ = ∣ 36 − 18 − 14 ∣ = 4 ✓

x = −3: ∣ (−3)2 − 3(−3) − 14 ∣ = ∣ 9 + 9 − 14 ∣ = 4 ✓

x = 5: ∣ 52 − 3(5) − 14 ∣ = ∣ 25 − 15 − 14 ∣ = 4 ✓

x = −2: ∣ (−2)2 − 3(−2) − 14 ∣ = ∣ 4 + 6 − 14 ∣ = 4 ✓

So, the solutions are x = −3, x = −2, x = 5, and x = 6.

b. x2 = ∣ x ∣ + 6 x2 = x + 6 or x2 = −x + 6 x2 − x − 6 = 0 or x2 + x − 6 = 0 (x − 3)(x + 2) = 0 or (x + 3)(x − 2) = 0 x − 3 = 0 or x + 2 = 0 or x + 3 = 0 or x − 2 = 0 x = 3 or x = −2 or x = −3 or x = 2 Check:

x = 3: 32 =?

∣ 3 ∣ + 6

9 =?

3 + 6

9 = 9 ✓

x = −2: (−2)2 =?

∣ −2 ∣ + 6

4 =?

2 + 6 4 ≠ 8 ✗

x = −3: (−3)2 =?

∣ −3 ∣ + 6

9 =?

3 + 6

9 = 9 ✓

x = 2: 22 =?

∣ 2 ∣ + 6

4 =?

2 + 6 4 ≠ 8 ✗

So, the solutions are x = −3 and x = 3.


Chapter 3

72. The Quadratic Formula will be more effi cient because

completing the square will require factoring out a 4 from the

fi rst two terms, resulting in a decimal coeffi cient with the

x-term.

73. The solutions of the equation ax2 + bx + c = 0 are

x = −b − √—

b2 − 4ac ——

2a and x = −b + √

— b2 − 4ac ——

2a . Then the

average of the two solutions is

−b − √

— b2 − 4ac ——

2a + −b + √

— b2 − 4ac ——

2a ————

2

= ( −b − √

— b2 − 4ac ) + ( −b + √

— b2 − 4ac ) ————

2a

———— 2

= −b − √

— b2 − 4ac + ( −b ) + √

— b2 − 4ac ————

2a

———— 2

= −2b

— 2a

—

2

= − b — 2a

.

The average of the x-intercepts of a parabola is the

x-coordinate of the vertex of the parabola, so the average of

the solutions is the same value as the axis of symmetry of

the parabola because the axis of symmetry is given by the

x-coordinate of the vertex.

74. Sample answer: You are walking down the street while

tossing a baseball in the air and catching it. You toss the

ball with an initial velocity of 10 feet per second from the

height of 3 feet and catch it again at the same height. You

miss the ball and it lands on the ground, how long was the

ball in the air? The equation that models the height is

h = −16t2 + 10t + 3. To fi nd how long the ball was in the

air, solve for t when h = 0.

h = −16t2 + 10t + 3 0 = −16t2 + 10t + 3

t = −10 ± √——

102 − 4(−16)(3) ——— −16(2)

t = −10 ± √

— 292 ——

−32

t ≈ −0.22 or t ≈ 0.85

Reject the negative solution, −0.22, because the time in the

air cannot be negative. So, the ball was in the air for about

0.85 second.

75. Suppose the solutions of the equation ax2 + bx + c = 0

are x = 3i and x = −2i. So, x − 3i = 0 and x + 2i = 0. It

follows that

a(x − 3i)(x + 2i) = 0 a[x2 − ix + (−6)i2] = 0 ax2 − aix + 6a = 0 a and ai cannot both be real numbers. So, there are no real

numbers a, b, and c such that ax2 + bx + c = 0 has solutions

x = 3i and x = −2i.

76. a. Because the maximum height occurs at the vertex of the

parabola given by the function, fi rst fi nd the vertex of the

parabola. The t-coordinate of the vertex is

t = − v0 —

2(−16) =

v0 —

32 .

The h-coordinate of the vertex is

h = −16 ( v0 — 32

) 2 + v0 ( v0 — 32

) + 921

= − v

0 2 ___

64 +

v 0 2 ___

32 + 921

= v

0 2 ___

64 + 921

To fi nd the initial velocity solve for v0 when h = 1081.

1081 = v

0 2 ___

64 + 921

160 = v

0 2 ___

64

10,240 = v 0 2

± √—

10,240 = v0

±101.19 ≈ v0

Reject the negative solution, −101.19, because the initial

velocity cannot be negative because the object is going up.

So, the initial velocity is about 101.19 feet per second.

b. The equation that models the height of the ride is

h = −16t2 + 32 √—

10 t + 921. The maximum height is

1081 feet. To fi nd the time it takes to reach the maximum

height, solve for t when h = 1081.

h = −16t2 + 32 √—

10 t + 921

1081 = −16t2 + 32 √—

10 t + 921

0 = −16t2 + 32 √—

10 t − 160

t = −32 √—

10 ± √———

(32 √—

10 )2 − 4(−16)(−160) ————

2(−16)

t = −32 √—

10 ± √—

0 —— −32

t = √—

10 ≈ 3.2

So, according to the model, it takes about 3.2 seconds

to ride up the needle. If you substitute t = 2 into the

equation, you get h ≈ 1059. This is close to the maximum

height, so the model is accurate.


Chapter 3


77.

x

y

4

2

4 62

(4, 5)

The solution is (4, 5).

78.

x

y

4

6

2

42

(2, 3)


79.

x

y

4

−4

−2

The graphs do not intersect, so there is no solution.

80.

x

y

4

−4

2−2

(0, 2)


81. y

4 62−2−4

−4

−6

−8

4

2

x

(1, 2)

x = 1

y = −x2 + 2x + 1

82.

42−2−4

6

8

2

x

y

(0.25, 2.875)

x = 0.25

y = 2x2 − x + 3

83.

42−4−6

6

8

2

x

10y

−2

(−2, 3)

x = −2

y = 0.5x2 + 2x + 5

84.

42−4 −2

2

x

y

−2

−6

−8

−10

(0, −2)

x = 0

y = −3x2 − 2

3.5 Explorations (p.131)

1. a. A; The graph shows a linear function with a positive

slope and y-intercept, and a parabola with the vertex

at the origin; (−1, 1), (2, 4)

b. E; The graph shows a linear function with a positive

slope and y-intercept, and a parabola with the vertex at

x = − 1 — 2 ; (−2, 0), (2, 4)

c. F; The graph shows a linear function with a negative slope

and a positive y-intercept, and a parabola with vertex at

x = 1; (−2, 3), (3, −2)

d. B; The graph shows two parabolas, each with vertex

at x = −1; (−3, 0), (2, 0)

e. C; The graph shows two parabolas, each with vertex

at x = 1; (1, 0)

f. D; The graph shows two parabolas, one with vertex at

x = −1, and the other with vertex at x = 1; (−1, 0), ( 1 — 2 ,

9 —

4 )


Chapter 3

2. a. x y = x2 + 2x + 1 y = −x2 + x + 2

−2 1 −4

−1.5 0.25 −1.75

−1 0 0

−0.5 0.25 1.25

0 1 2

0.5 2.25 2.25

1 4 2

1.5 6.25 1.25

2 9 0

The solutions of the system are (−1, 0) and (0.5, 2.25).

b. Set the two equations equal to each other, then solve for x.

x2 + 2x + 1 = −x2 + x + 2 2x2 + x − 1 = 0 (2x − 1)(x + 1) = 0 2x − 1 = 0 or x + 1 = 0

x = 1 — 2 or x = −1

Substitute these values into the fi rst equation.

y = ( 1 — 2 )

2

− 2 ( 1 — 2 ) + 1 = 9 —

4

y = (−1)2 − 2(−1) + 1 = 0

So, the solutions are ( 1 — 2 ,

9 —

4 ) and (−1, 0).

3. To solve a nonlinear system of equations, use graphs, the

elimination method, the substitution method, or a table or

spreadsheet.

4. Sample answer: Use the analytical approach to solve the

system. This method will result in an accurate solution that

the graphical and numerical approaches may not produce.

3.5 Monitoring Progress (pp.133–135)

1. Substitute −4x + 8 for y in Equation 1 and solve for x.

−x2 + 4 = −4x + 8 −x2 + 4x − 4 = 0 x2 − 4x + 4 = 0 (x − 2)2 = 0 x − 2 = 0 x = 2 To solve for y, substitute x = 2 into the equation

y = −4x + 8.

y = −4x + 8 = −4(2) + 8 = 0 The solution is (2, 0).

2. Begin by solving for y in both equations.

y = −x2 − 3x Equation 1

y = −2x + 5 Equation 2

Next, substitute −2x + 5 for y in Equation 1 and solve for x.

−x2 − 3x = −2x + 5 0 = x2 + x + 5

x = −1 ± √——

(1)2 − 4(1)(5) ——

2(1)

x = −1 ± √—

−19 ——

2

x = −1 ± i √—

19 —

2

Because x is an imaginary number, the system does not have

a real solution.

3. Begin by solving for y in both equations.

y = 2x2 + 4x + 2 Equation 1

y = 2 − x2 Equation 2

Next, substitute 2 − x2 for y in Equation 1 and solve for x.

2 − x2 = 2x2 + 4x + 2 0 = 3x2 + 4x

0 = x(3x + 4)

x = 0 or 3x + 4 = 0

x = 0 or x = − 4 — 3

To solve for y, substitute x = 0 and x = − 4 — 3 into the

equation y = 2 − x2.

y = 2 − x2 = 2 − 02 = 2

y = 2 − x2 = 2 − ( − 4 — 3 )

2

= 2 — 9

The solutions are (0, 2) and ( − 4 — 3 ,

2 —

9 ) .

4. Substitute −x + 4 for y in Equation 1 and solve for x.

x2 + (−x + 4)2 = 16

x2 + x2 − 8x + 16 = 16

2x2 − 8x = 0 2x(x − 4) = 0 2x = 0 or x − 4 = 0 x = 0 or x = 4 To solve for y, substitute x = 0 and x = 4 into the equation

y = −x + 4.

y = −x + 4 = −0 + 4 = 4 y = −x + 4 = −4 + 4 = 0 The solutions are (0, 4) and (4, 0).


Chapter 3

5. Substitute x + 4 for y in Equation 1 and solve for x.

x2 + (x + 4)2 = 4 x2 + x2 + 8x + 16 = 4 2x2 + 8x + 12 = 0 x2 + 4x + 6 = 0

x = −4 ± √——

42 − 4(1)(6) ——

2(1)

x = −4 ± √—

−8 —

2

x = −4 ± 2i √—

2 —

2

x = −2 ± i √—

2


a real solution.

6. Substitute 1 —

2 x + 1 —

2 for y in Equation 1 and solve for x.

x2 + ( 1 — 2 x + 1 —

2 )

2

= 1

x2 + 1 — 4 x2 + 1 —

2 x + 1 —

4 = 1

5 —

4 x2 + 1 —

2 x − 3 —

4 = 0

5x2 + 2x − 3 = 0 (5x − 3)(x + 1) = 0 5x − 3 = 0 or x + 1 = 0

x = 3 — 5 or x = −1

To solve for y, substitute x = 3 — 5 and x = −1 into the equation

y = 1 — 2 x + 1 — 2 .

y = 1 — 2 x + 1 —

2 = 1 —

2 ( 3 — 5 ) + 1 —

2 = 3 —

10 + 1 —

2 = 4 —

5

y = 1 — 2 x + 1 —

2 = 1 —

2 (−1) + 1 —

2 = − 1 —

2 + 1 —

2 = 0

The solutions are ( 3 — 5 ,

4 —

5 ) and (−1, 0)

7.

x

y

4

6

8

2

42

(3, 6)

The solution is x = 3.

8.

x

y8

4

8−8

(2, 6)

(−2, −6)


3.5 Exercises (pp.136–138)


1. The possible number of solutions of a system of two

quadratic equations is two, when they intersect in two points;

one, when they intersect in one point; or zero, when they do

not intersect.

2. The second system of equations, y = 2x − 1 and

y = −3x + 6, does not belong because it is a system of linear

equations and the other systems include second-degree

equations.


3.

x

y4

2−4(−2, 0)

(0, 2)

The solutions are (−2, 0) and (0, 2).

Check:

x = −2:

y = x + 2 = (−2) + 2 = 0 y = 0.5(x + 2)2 = 0.5(−2 + 2)2 = 0 x = 0:

y = x + 2 = 0 + 2 = 2 y = 0.5(x + 2)2 = 0.5(0 + 2)2 = 2

4.

x

y

4

6

8

2

4 62

(3, 5)


Check:

x = 3:

y = (x − 3)2 + 5 = (3 − 3)2 + 5 = 5 y = 5


Chapter 3

5.

x

y

4−4

4


6.

x

y4

−8

−4

−8−12

(−6, 1)

(−3, −8)

The solutions are (−6, 1) and (−3, −8).

Check:

x = −6:

y = −3x2 − 30x − 71 = −3(−6)2 − 30(−6) − 71 = 1 y = −3x − 17 = −3(−6) − 17 = 1

x = −3:

y = −3x2 − 30x − 71 = −3(−3)2 − 30(−3) − 71 = −8

y = −3x − 17 = −3(−3) − 17 = −8

7.

x

y

4

2

−2−4−6

(−4, 2)

The solution is (−4, 2).

Check:

x = −4:

y = x2 + 8x + 18 = (−4)2 + 8(−4) + 18 = 2 y = −2x2 − 16x − 30 = −2(−4)2 − 16(−4) − 30 = 2

8. x

y

−8

−4

4−4


9.

x

y4

2

−2

42

(1, 1)

(3, 1)

The solutions are (1, 1) and (3, 1).

Check:

x = 1:

y = (x − 2)2 = (1 − 2)2 = 1 y = −x2 + 4x − 2 = −(1)2 + 4(1) − 2 = 1

x = 3:

y = (x − 2)2 = (3 − 2)2 = 1 y = −x2 + 4x − 2 = −(3)2 + 4(3) − 2 = 1

10.

x

y4

−4

−4

(−2, 0)

(0, 2)


Check.

x = −2:

y = 1 — 2 (x + 2)2 = 1 —

2 (−2 + 2)2 = 0

y = − 1 — 2 x2 + 2 = − 1 —

2 (−2)2 + 2 = 0

x = 0:

y = 1 — 2 (x + 2)2 = 1 —

2 (0 + 2)2 = 2

y = − 1 — 2 x2 + 2 = − 1 —

2 (0)2 + 2 = 2

11. The solution is (−4, 1).

12. There is no solution because the graphs do not intersect.

13. The solutions are (1, 4) and (9, 4).

14. The solutions are (−4, 0) and (1, 5).


Chapter 3


x + 5 = x2 − x + 2 0 = x2 − 2x − 3 0 = (x − 3)(x + 1)

x − 3 = 0 or x + 1 = 0 x = 3 or x = −1

To solve for y, substitute x = 3 and x = −1 into the equation

y = x + 5.

y = x + 5 = 3 + 5 = 8 y = x + 5 = −1 + 5 = 4 The solutions are (3, 8) and (−1, 4).

16. Substitute 7 − x for y in Equation 1 and solve for x.

x2 + (7 − x)2 = 49

x2 + 49 − 14x + x2 = 49

2x2 − 14x + 49 = 49

2x2 − 14x = 0 2x(x − 7) = 0 2x = 0 or x − 7 = 0 x = 0 or x = 7 To solve for y, substitute x = 0 and x = 7 into the equation

y = 7 − x.

y = 7 − x = 7 − 0 = 7 y = 7 − x = 7 − 7 = 0 The solutions are (0, 7) and (7, 0).

17. Substitute −8 for y in Equation 1 and solve for x.

x2 + (−8)2 = 64

x2 + 64 = 64

x2 = 0 x = 0 The solution is (0, −8).

18. Substitute 3 for x in Equation 2 and solve for y.

−3(3)2 + 4(3) − y = 8 −15 − y = 8 −y = 23

y = −23

The solution is (3, −23).

19. Begin by solving for y in Equation 2.

y = 2x − 4 Next, substitute 2x − 4 for y in Equation 1 and solve for x.

2x2 + 4x − (2x − 4) = −3

2x2 + 2x + 4 = −3

2x2 + 2x + 7 = 0

x = −2 ± √——

22 − 4(2)(7) ——

2(2)

x = −2 ± √—

−52 ——

4

x = −2 ± 2i √—

13 ——

4


a real solution.

20. Substitute −3x − 3 for y in Equation 1 and solve for x.

2x − 3 = (−3x − 3) + 5x2

2x − 3 = −3x − 3 + 5x2

0 = 5x2 − 5x

0 = 5x(x − 1)

5x = 0 or x − 1 = 0 x = 0 or x = 1

To solve for y, substitute x = 0 and x = 1 into the equation

y = −3x − 3.

y = −3x − 3 = −3(0) − 3 = −3

y = −3x − 3 = −3(1) − 3 = −6

The solutions are (0, −3) and (1, −6).

21. Substitute x2 − 1 for y in Equation 2 and solve for x.

−7 = −x2 − (x2 − 1)

−7 = −x2 − x2 + 1 −8 = −2x2

4 = x2

±2 = x To solve for y, substitute x = −2 and x = 2 into the equation

y = x2 − 1.

y = x2 − 1 = (−2)2 − 1 = 3 y = x2 − 1 = (2)2 − 1 = 3 The solutions are (−2, 3) and (2, 3).


Chapter 3


y = 4x2 − 16x + 22

Next, substitute 4x2 − 16x + 22 for y in Equation 2 and

solve for x.

4x2 − 24x + 26 + 4x2 − 16x + 22 = 0 8x2 − 40x + 48 = 0 x2 − 5x + 6 = 0 (x − 3)(x − 2) = 0 x − 3 = 0 or x − 2 = 0 x = 3 or x = 2 To solve for y, substitute x = 3 and x = 2 into the equation

y = 4x2 − 16x + 22.

y = 4x2 − 16x + 22 = 4(3)2 − 16(3) + 22 = 10

y = 4x2 − 16x + 22 = 4(2)2 − 16(2) + 22 = 6 The solutions are (3, 10) and (2, 6).

23. Begin by solving for x in Equation 2.

x = 21 − 3y

Next, substitute 21 − 3y for x in Equation 1 and solve for y.

(21 − 3y)2 + y2 = 7 441 − 126y + 9y2 + y2 = 7 10y2 − 126y + 434 = 0 5y2 − 63y + 217 = 0

x = 63 ± √——

(−63)2 − 4(5)(217) ———

2(5)

x = 63 ± √—

−371 ——

10

x = 63 ± i √—

371 ——

10


a real solution.


y = −1 + x Next, substitute −1 + x for y in Equation 1 and solve for x.

x2 + (−1 + x)2 = 5 x2 + 1 − 2x + x2 = 5

2x2 − 2x − 4 = 0 x2 − x − 2 = 0 (x − 2)(x + 1) = 0 x − 2 = 0 or x + 1 = 0 x = 2 or x = −1

To solve for y, substitute x = 2 and x = −1 into the equation

y = −1 + x.

y = −1 + x = −1 + 2 = 1 y = −1 + x = −1 + (−1) = −2

The solutions are (2, 1) and (−1, −2).

25. A and C; Set the equations equal to each other and solve

for x.

1 —

2 x2− 5x + 21 ___

2 = −

1 —

2 x+

13 —

2

1 — 2 x2 − 9 __

2 x + 8 __

2 = 0

x2 − 9x + 8 = 0

(x − 1)(x − 8) = 0

x = 1 or x = 8


y = − 1 —

2 x +

13 —

2 .

y = − 1 —

2 x +

13 —

2 = −

1 —

2 (1) + 13 ___

2 = 6

y = − 1 —

2 x +

13 —

2 = −

1 —

2 (8) + 13 ___

2 = 5 __

2

The solutions are (1, 6) and (8, 2.5).

26. A; Set the equations equal to each other and solve for x.

7x2 − 11x + 9 = −7x2 + 5x − 3 14x2 − 16x + 12 = 0 7x2 − 8x + 6 = 0

x = 8 ± √——

(−8)2 − 4(7)(6) ——

2(7)

x = 8 ± √—

−104 —

14

x = 8 ± 2i √—

26 —

14


a real solution.

27. Add the equations to eliminate the y-term and obtain a

quadratic equation in x.

2x2 − 3x − y = −5

−x + y = 5

2x2 − 4x = 0 2x(x − 2) = 0

x = 0 or x = 2 To solve for y, substitute x = 0 and x = 2 into the equation

y = x + 5.

y = x + 5 = 0 + 5 = 5 y = x + 5 = 2 + 5 = 7 The solutions are (0, 5) and (2, 7).


Chapter 3



−3x2 + 2x − 5 = y

−x + 2 = −y

−3x2 + x − 3 = 0

x = −1 ± √

—— 12 − 4(−3)(−3) ———

2(−3)

x = −1 ± √

— −35 ——

−6

Because the discriminant is negative, the equation

−3x2 + x −3 = 0 has no real solution. So, the original

system has no real solution.



−3x2 + y = −18x + 29

−3x2 − y = 18x − 25

−6x2 = 4

x = ±i √

— 6 —

3

Because x is an imaginary number, the system has no real

solution.

30. Subtract the equations to eliminate the y-term and obtain a


y = −x2 − 6x − 10

−( y = 3x2 + 18x + 22)

0 = −4x2 − 24x − 32

0 = −4(x + 2)(x + 4)

0 = (x + 2) or 0 = (x + 4)

x = −2 or x = −4

To solve for y, substitute x = −2 and x = −4 into the

equation y = −x2 − 6x − 10.

y = −x2 − 6x − 10 = −(−2)2 − 6(−2) − 10 = −2

y = −x2 − 6x − 10 = −(−4)2 − 6(−4) − 10 = −2

The solutions are (−2, −2) and (−4, −2).



y + 2x = −14

−x2 − y − 6x = 11

−x2 − 4x = −3

−x2 − 4x + 3 = 0

x = −(−4) ± √

—— (−4)2 − 4(−1)(3) ———

2(−1)

x = 4 ± √

— 28 —

−2

x = 4 ± 2 √

— 7 —

−2

x = −2 ± √—

7

To solve for y, substitute x = −2 − √—

7 and x = −2 + √—

7

into the equation y = −2x − 14.

y = −2x − 14 = −2(−2 − √—

7 ) − 14 = −10 + 2 √—

7

y = −2x − 14 = −2(−2 + √—

7 ) − 14 = −10 − 2 √—

7

The solutions are (−2 − √—

7 , −10 + 2 √—

7 ) and

(−2 + √—

7 , −10 − 2 √—

7 ).



y = x2 + 4x + 7

−y = 4x + 7

0 = x2 + 8x + 14

x = −8 ± √

—— 82 − 4(1)(14) ——

2(1)

x = −8 ± √

— 8 —

2

x = −8 ± 2 √

— 2 —

2

x = −4 ± √—

2

To solve for y, substitute x = −4 − √—

2 and x = −4 + √—

2

into the equation y = −4x − 7.

y = −4x − 7 = −4(−4 − √—

2 ) − 7 = 9 + 4 √—

2

y = −4x − 7 = −4(−4 + √—

2 ) − 7 = 9 − 4 √—

2


2 , 9 + 4 √—

2 ) and

(−4 + √—

2 , 9 − 4 √—

2 ).


Chapter 3



y = −3x2 − 30x − 76

−(y = 2x2 + 20x + 44)

0 = −5x2 − 50x − 120

0 = x2 + 10x + 24

(x + 6)(x + 4) = 0

x + 6 = 0 or x + 4 = 0

x = −6 or x = −4

To solve for y, substitute x = −6 and x = −4 into the

equation y = 2x2 + 20x + 44.

y = 2x2 + 20x + 44 = 2(−6)2 + 20(−6) + 44 = −4

y = 2x2 + 20x + 44 = 2(−4)2 + 20(−4) + 44 = −4

The solutions are (−6, −4) and (−4, −4).



−10x2 + y = −80x + 155

−(5x2 + y = 40x − 85)

−15x2 = −120x + 240

0 = 15x2 − 120x + 240

0 = x2 − 8x + 16

0 = (x − 4)2

0 = x − 4

x = 4

To solve for y, substitute x = 4 into the equation

y = −5x2 + 40x − 85.

y = −5x2 + 40x − 85 = −5(4)2 + 40(4) − 85 = −5


35. The terms that were added were not like terms.

y = −2x2 + 32x − 126

−y = 2x − 14

0 = −2x2 + 34x − 140

x2 − 17x + 70 = 0

(x − 7)(x − 10) = 0

x − 7 = 0 or x − 10 = 0

x = 7 or x = 10

36. The solutions are (−1, 9) and (7, 9) because the x-values

have the same y-value.

37. Use the elimination method because there is a y and a −y on

the left-hand sides of the equations.

Add the equations to eliminate the y-term and obtain a


y = x2 − 1

−y = 2x2 + 1

0 = 3x2

0 = x

To solve for y, substitute x = 0 into the equation y = x2 − 1.

y = x2 − 1 = 02 − 1 = −1


38. Use the substitution method because one equation is solved

for y.

Substitute −4x2 − 16x − 13 for y in Equation 2 and solve

for x.

−3x2 + (−4x2 − 16x − 13) + 12x = 17

−7x2 − 4x − 13 = 17

−7x2 − 4x − 30 = 0

x = 4 ± √

— −824 —

−14

Because the discriminant is negative, the equation

−7x2 − 4x − 30 = 0 has no real solution. So, the original

system has no real solution.


for y.

Substitute 10 for y in Equation 1 and solve for x.

−2x + 10 + 10 = 1 —

3 x2

0 = 1 —

3 x2 + 2x − 20

0 = x 2 + 6x − 60

x = −3 ± √—

69


69 , 10) and (−3 + √—

69 , 10).

40. Use the elimination method because there is a y on the

left-hand side of each equation.

Subtract the equations to eliminate the y-term and obtain a


y = 0.5x2 − 10

−( y = −x2 + 14)

0 = 1.5x2 − 24

24 = 1.5x2

16 = x2

±4 = x

To solve for y, substitute x = −4 and x = 4 into the equation

y = −x2 + 14.

y = −x2 + 14 = −(−4)2 + 14 = −2

y = −x2 + 14 = −(4)2 + 14 = −2

The solutions are (−4, −2) and (4, −2).


Chapter 3


for y.

Substitute −3(x − 4)2 + 6 for y in Equation 2 and solve for x.

(x − 4)2 + 2 − [ −3(x − 4)2 + 6 ] = 0

(x − 4)2 + 2 + 3 (x − 4)2 − 6 = 0

4(x − 4)2 − 4 = 0

(x − 4)2 = 1

x − 4 = ±1

x = 4 ± 1

x = 3 or x = 5


y = −3(x − 4)2 + 6.

y = −3(x − 4)2 + 6 = −3(3 − 4)2 + 6 = 3

y = −3(x − 4)2 + 6 = −3(5 − 4)2 + 6 = 3

The solutions are (3, 3) and (5, 3).


for y.

Substitute −x + 14 for y in Equation 1 and solve for x.

−x2 + (−x + 14)2 = 100

−x2 + x2 − 28x + 196 = 100

−28x + 196 = 100

−28x = −96

x = 24

— 7

To solve for y, substitute x = 24

— 7 into the equation y = −x + 14.

y = −x + 14 = − ( 24 —

7 ) + 14 =

74 —

7

The solution is ( 24 —

7 ,

74 —

7 ) .

43. Write a system of equations using each side of the original

equation.

Equation System

x2 + 2x = − 1 — 2 x2 + 2x y = x2 + 2x

y = − 1 — 2 x2 + 2x

Graph the equations in the same plane.

x

y3

1

3(0, 0)−3

The solution is x = 0.


equation.

Equation System

2x2 − 12x − 16 = y = 2x2 − 12x − 16

−6x2 + 60x − 144 y = −6x2 + 60x − 144


x

y

−10

10 20 30−10

−20

−30(2.44, −33.37)

(6.56, −8.65)

The solutions are x ≈ 2.44 and x ≈ 6.56.


equation.

Equation System

(x + 2)(x − 2) = −x2 + 6x − 7 y = (x + 2)(x − 2)

y = −x2 + 6x − 7


x

y

2

−4

8 124−4

(0.63, −3.60)

(2.37, 1.62)

The solutions are x ≈ 0.63 and x ≈ 2.37.


equation.

Equation System

−2x2 − 16x − 25 = 6x2 + 48x + 95 y = −2x2 − 16x − 25

y = 6x2 + 48x + 95


x

y

3

7

1

−1 1−5−7

(−5, 5) (−3, 5)



Chapter 3


equation.

Equation System

(x − 2)2 − 3 = y = (x − 2)2 − 3

(x + 3)(−x + 9) − 38 y = (x + 3)(−x + 9) − 38


x

y

−4

−2

42

(3, −2)

(2, −3)

The solutions are x = 2 and x = 3.


equation.

Equation System

(−x + 4)(x + 8) − 42 = y = (−x + 4)(x + 8) − 42

(x + 3)(x + 1) − 1 y = (x + 3)(x + 1) − 1


x

y

−4

−6

−2

−2−4


49. Because there is only one solution to the system, the graph

of the constant function intersects the graph of the quadratic

function at the vertex.

50. To fi nd when the car is able to receive a broadcast from the

radio tower, fi nd the solution to the nonlinear system:

x2 + y2 = 1620 Equation 1

y = − 1 — 3 x + 30 Equation 2

Substitute − 1 —

3 x + 30 for y in Equation 1 and solve for x.

x2 + ( − 1 — 3 x + 30 )

2

= 1620

x2 + 1 —

9 x2 − 20x + 900 = 1620

10

— 9 x2 − 20x − 720 = 0

x2 − 18x − 648 = 0

(x + 18)(x − 36) = 0

x + 18 = 0 or x − 36 = 0

x = −18 or x = 36

To solve for y, substitute x = −18 and x = 36 into the

equation y = − 1 — 3 x + 30.

y = − 1 — 3 x + 30 = − 1 —

3 (−18) + 30 = 36

y = − 1 — 3 x + 30 = − 1 —

3 (36) + 30 = 18

The solutions are (−18, 36) and (36, 18). Use the Distance

Formula to fi nd the length of highway that cars are able to

receive the broadcast signal.

d = √———

(36 − (−18))2 + (18 − 36)2

= √——

542 + (−18)2

= √—

3240 ≈ 56.9

So, cars can receive the broadcast signal for about 56.9 miles

of the highway.

51. Because the rate of the car is 0.8 mile per minute, the

equation that gives the distance d (in miles) it travels in t minutes is d = 0.8t. The distance d (in miles) the police car

will travel in t minutes is given by the equation d = 2.5t2. Because the police car starts to move when the car passes, to

fi nd the amount of time needed for the police car to catch up

to the car, solve the system:

d = 0.8t Equation 1

d = 2.5t2 Equation 2

Substitute 0.8t for d in Equation 2 and solve for t.

0.8t = 2.5t2

0 = 2.5t2 − 0.8t

0 = t(2.5t − 0.8)

t = 0 or t = 0.32

Reject the solution t = 0 because time must be greater than

zero. So, it takes 0.32 minute for the police car to catch up

with the car that passed it.


Chapter 3

52. Sample answer: The system

y = x2

y = 1

has two solutions with the same y-coordinate.

x

y

4

6

2

42−2−4

(1, 1)(−1, 1)

Substitute 1 for y in Equation 1 and solve for x.

x2 = 1

x = ±1


53. When m = 1 the system has no solution, when m = 0 there

is one solution, and when m = −1 there are two solutions.

y

4 6 8 102−2−4

−6

−8

−10

−2

2

4

6

8

10

x

y = x + 3

y = −x + 3

y = 3

y = −13x2 + 83x − 73

54. Your friend is correct because Equation 1, when graphed,

forms a circle and Equation 2, when graphed, forms a line

and a line can intersect a circle in at most two points.

55. Sample answer: To solve the equation

−2x2 + 12x − 17 = 2x2 − 16x + 31, you can rewrite the

equation with a zero on one side of the equation and use the

Quadratic Formula. Another way would be to set the left

and right hand sides equal to y, graph the new equations,

and fi nd where they intersect. The preferred way is to solve

analytically because graphing may be inaccurate.

56. To fi nd the other point of intersection, use the known point

and refl ect the point in the origin. So, if (x, y) is a solution,

then (−x, −y) must be a solution.

57. a. The graphs of a quadratic equation and a circle can

intersect in 0 points, 1 point, 2 points, 3 points, or

4 points. So, the system of a quadratic equation and a

circle will have no solution, one solution, two solutions,

three solutions, or four solutions.

y

42−2−4

−4

−2

2

6

y

42−2−4

−4

−2

2

4

6

x

y

42−2−4

−4

−2

2

4

x

y

42−2−4

−4

−2

2

4

6

x

y

4−4

−4

−2

2

4

x


Chapter 3

b. The graphs of two circles can intersect in 0 points, 1 point,

2 points, or infi nitely many points. So, the system of two

circles will have no solution, one solution, two solutions,

or infi nitely many solutions.

y

2−2−6−8

−2

2

x

y

4 62

4

6

x

y

4−2

−2

2

4

x

y

42−2−4

−4

−2

2

4

x

58. The solutions are (0, 0) and (2, 3). To have a system that has

no solution, translate the linear function up, for example, 5

units. After the translation, the two graphs will not intersect.

59. a. The equation that represents the circle is x2 + y2 = 1. The

slope of the line that represents Oak Lane is m = − 1 — 7 with

the line passing through the point (5, 0). So, the equation

that represents Oak Lane is y = − 1 — 7 x +

5 —

7 .

b. Use the equations in part (a) to form a system of equations.

x2 + y2 = 1 Equation 1

y = − 1 — 7 x +

5 —

7 Equation 2

Substitute − 1 — 7 x +

5 —

7 for y in Equation 1 and solve for x.

x2 + ( − 1 — 7 x +

5 —

7 )

2

= 1

x2 + 1 —

49 x2 −

10 —

49 x +

25 —

49 = 1

50

— 49

x2 − 10

— 49

x − 24

— 49

= 0

50x2 − 10x − 24 = 0

x = − 3 — 5 or x =

4 —

5

To solve for y, substitute x = − 3 — 5 and x = 4 —

5 into

y = − 1 — 7 x + 5 —

7 and solve for y.

y = − 1 — 7 x +

5 —

7 = − 1 —

7 ( − 3 —

5 ) +

5 —

7 =

4 —

5

y = − 1 — 7 x +

5 —

7 = − 1 —

7 ( 4 —

5 ) +

5 —

7 =

3 —

5

The solutions are ( − 3 — 5 , 4 —

5 ) and ( 4 —

5 ,

3 —

5 ) .

c. The length of Oak Lane where students are not eligible

for parking passes is the same as the distance between the

points ( − 3 —

5 ,

4 —

5 ) and ( 4 —

5 ,

3 —

5 ) . The distance between ( −

3 —

5 ,

4 —

5 )

and ( 4 — 5 ,

3 —

5 ) is d = √——

( − 3 —

5 −

4 —

5 )

2

+ ( 4 — 5 −

3 —

5 )

2

= √—

2 ≈ 1.41.

So, there are about 1.41 miles of Oak Lane where students

are not eligible for parking passes.


Chapter 3

60. Substitute −x + 2 for y in Equation 2 and solve for x.

2(−x + 2) = x2 − 2x + 4

−2x + 4 = x2 − 2x + 4

0 = x2

0 = x

To solve for y, substitute x = 0 into y = −x + 2 and solve

for y.

y = −x + 2 = −0 + 2 = 2

The solution is (0, 2). Check that the solution works in

x2 + y2 = 4 by substituting x = 0 and y = 2 into x2 + y2 = 4.

x2 + y2 = 4

02 + 22 = 4

4 = 4

The equation is valid, so the point works as a solution for all

three equations.


61. 4x − 4 > 8

4x > 12

x > 3

1−1 3 5 7

62. −x + 7 ≤ 4 − 2x

x ≤ −3

−3 −1−7 −5 1

63. −3(x − 4) ≥ 24

−3x + 12 ≥ 24

−3x ≥ 12

x ≤ −4

−4 −2 0−8 −6

64. The equation of the line is y = −x + 1. Because the line is

solid and the plane is shaded above the line, the inequality

is y ≥ −x + 1.

65. The equation of the line is y = x − 2. Because the line is

dashed and the plane is shaded below the line, the inequality

is y < x − 2.

66. The equation of the line is y = 2x − 3. Because the line is

dashed and the plane is shaded above the line, the inequality

is y > 2x − 3.


1. You can use the graph of the function to solve the inequality

by fi nding on what interval of x-values the graph of the

function touches or is below the x-axis. The solution to the

inequality is −3 ≤ x ≤ 1.

2. a. A; The x-coordinate of the vertex of the graph is − b — 2a

= 3 —

2 .

The y-coordinate is ( 3 — 2 )

2

−3 ( 3 — 2 ) + 2 = − 1 —

4 . So, the vertex

is ( 3 — 2 , − 1 —

4 ) . The y-intercept is (0, 2). The x-intercepts are

(1, 0) and (2, 0). The graph is above the x-axis for

x < 1 or x > 2.

b. C; The x-coordinate of the vertex of the graph is − b — 2a

= 2.

The y-coordinate is 22 − 4(2) + 3 = −1. So, the vertex

is (2, −1). The y-intercept is (0, 3). The x-intercepts are

(1, 0) and (3, 0). The graph touches or is below the x-axis

for 1 ≤ x ≤ 3.

c. E; The x-coordinate of the vertex of the graph is

− b — 2a

= 1. The y-coordinate is 12 − 2(1) − 3 = −4.

So, the vertex is (1, −4). The y-intercept is (0, −3). The

x-intercepts are (−1, 0) and (3, 0). The graph is below the

x-axis for −1 < x < 3.

d. B; The x-coordinate of the vertex of the graph is

− b — 2a

= − 1 — 2 . The y-coordinate is

( − 1 — 2 )

2

+ ( − 1 — 2 ) − 2 = − 9 —

4 . So, the vertex is ( − 1 —

2 , − 9 —

4 ) .

The y-intercept is (0, −2). The x-intercepts are (−2, 0)

and (1, 0). The graph touches or is above the x-axis for

x ≤ −2 or x ≥ 1.

e. F; The x-coordinate of the vertex of the graph is − b — 2a

= 1 —

2 .

The y-coordinate is ( 1 — 2 )

2

− 1 — 2 − 2 = − 9 —

4 . So, the vertex is

( 1 — 2 , − 9 —

4 ) . The y-intercept is (0, −2). The x-intercepts are

(−1, 0) and (2, 0).The graph is below the x-axis for

−1 < x < 2.

f. D; The x-coordinate of the vertex of the graph is

− b — 2a

= 0. The y-coordinate is 02 − 4 = −4. So, the

vertex is (0, −4). The y-intercept is (0, −4). The

x-intercepts are (−2, 0) and (2, 0). The graph is above the

x-axis for x < −2 or x > 2.

3. To solve a quadratic inequality, rewrite the inequality so that

one side is 0. Graph the related quadratic function to see

where the graph is above or below the x-axis.

4. To use a graph to solve a quadratic inequality, graph the

related quadratic equation, and then fi nd the intervals on the

x-axis where the inequality is satisfi ed.

a. The solution of the inequality is x < −3 or x > 1.

b. The solution of the inequality is −3 < x < 1.

c. The solution of the inequality is x ≤ −3 or x ≥ 1.


Chapter 3


1. Step 1 Graph y = x2 + 2x − 8. Because the inequality

symbol is ≥, make the parabola solid.

Step 2 Test a point inside the parabola, such as (0, 0).

y ≥ x2 + 2x − 8

0 ≥? 02 + 2(0) − 8

0 ≥ −8

So, (0, 0) is a solution of the inequality.

Step 3 Shade the region inside the parabola.

y

−2

−4

−6

−8

−10

−2

2

x

2. Step 1 Graph y = 2x2 − x − 1. Because the inequality

symbol is ≤, make the parabola solid.

Step 2 Test a point outside the parabola, such as (2, 0).

y ≤ 2x2 − x − 1

0 ≤? 2(2)2 − 2 − 1

0 ≤ 5


Step 3 Shade the region outside the parabola.

y

42−2−4

2

4

6

x

3. Step 1 Graph y = −x2 + 2x + 4. Because the inequality

symbol is >, make the parabola dashed.


y > −x2 + 2x + 4

5 >? −02 + 2(0) + 4

5 > 4



y

4 62−2−4

−4

−6

−2

2

4

6

x

4. Step 1 Graph y ≤ −x2.

Step 2 Graph y > x2 − 3.

Step 3 Identify the region where the two graphs overlap.

This region is the graph of the system.

y

42−2−4

−4

−6

2

4

6

x

y > x2 − 3

y ≤ −x2


Chapter 3

5. First, write and solve the equation obtained by replacing ≤

with =.

2x2 + 3x = 2

2x2 + 3x − 2 = 0

(2x − 1) (x + 2) = 0

x = 1 —

2 or x = −2

The numbers 1 —

2 and −2 are critical values of the original

inequality. Plot 1 —

2 and −2 on a number line, using closed

dots because the values do satisfy the inequality. The critical

x-values partition the number line into three intervals. Test an

x-value in each interval to determine whether it satisfi es the

inequality.

Test x = 0.

Test x = −3.

6543210

Test x = 1.

−1−2−3

12

2(−3)2 + 3(−3) ≰ 2

2(0)2 + 3(0) ≤ 2 ✓

2(1)2 + 3(1) ≰ 2

So, the solution is −2 ≤ x ≤ 1 __

2 .

6. First, write and solve the equation obtained by replacing

< with =.

−3x2 − 4x + 1 = 0

x = 4 ± √

— 28 —

−6

x = −2 ± √

— 7 —

3

The numbers −2 − √

— 7 —

3 ≈ −1.55 and

−2 + √—

7 —

3 ≈ 0.22

are critical values of the original inequality. Plot −2 − √

— 7 —

3

and −2 + √

— 7 —

3 on a number line, using open dots because

the values do not satisfy the inequality. The critical x-values

partition the number line into three intervals. Test an x-value in

each interval to determine whether it satisfi es the inequality.

Test x = 0.Test x = −2.

210

Test x = 1.

−1−2−3

−2 − 73

−2 + 73

−3(−2)2 − 4(−2) + 1 < 0 ✓

−3(0)2 − 4(0) + 1 ≮ 0

−3(1)2 − 4(1) + 1 < 0 ✓

So, the solution is x < −2 − √

— 7 —

3 and x >

−2 + √—

7 —

3 .


> with = .

2x2 + 2 = −5x

2x2 + 5x + 2 = 0

(x + 2)(2x + 1) = 0

x = −2 or x = − 1 — 2

The numbers − 1 — 2 and −2 are critical values of the original

inequality. Plot − 1 — 2 and −2 on a number line, using open

dots because the values do not satisfy the inequality. The

critical x-values partition the number line into three intervals.

Test an x-value in each interval to determine whether it

satisfi es the inequality.

−

Test x = −1.

Test x = −3. 12

0 1 2−1−2−3−4

Test x = 0.

2(−3)2 + 2 > −5(−3) ✓

2(−1)2 + 2 ≯ −5(−1)

2(0)2 + 2 > −5(0) ✓

So, the solution is x < −2 and x > − 1 — 2 .

8. Let ℓ represent the length (in feet) and w represent the width

(in feet) of the parking lot.

Perimeter = 440 Area ≥ 8500

2ℓ + 2w = 440 ℓw ≥ 8500

Solve the perimeter equation for w to obtain w = 220 − ℓ.

Substitute this into the area inequality to obtain a quadratic

inequality in one variable.

ℓw ≥ 8500

ℓ(220 − ℓ) ≥ 8500

220ℓ − ℓ2 ≥ 8500

−ℓ2 + 220ℓ − 8500 ≥ 0

Use a graphing calculator to fi nd the ℓ-intercepts of

y = −ℓ2 + 220ℓ − 8500.

220

−5000

0

5000

The ℓ-intercepts are ℓ = 50 and ℓ = 170. The solution

consists of the ℓ-values for which the graph lies on or above

the ℓ-axis. The graph lies on or above the ℓ-axis when

50 ≤ ℓ ≤ 170. So, the length of the parking lot is at least

50 feet and at most 170 feet.


Chapter 3



1. The graph of a quadratic inequality in one variable is an

interval on the real number line, whereas the graph of a

quadratic inequality in two variables is a region in the

coordinate plane.

2. To solve the inequality algebraically, solve the related

quadratic equation x2 + 6x − 8 = 0 to fi nd the critical

values, x ≈ 1.12 and x ≈ −7.12. Partition the number line

into three intervals. Choose a value within each interval to

test the inequality. To solve the inequality graphically, graph

the related quadratic equation, and determine for which

intervals the graph is below the x-axis.


3. C; The x-intercepts are x = −1 and x = −3. The test point

(−2, 5) does not satisfy the inequality.

4. A; The x-intercepts are x = 1 and x = 3. The test point (2, 5)


5. B; The x-intercepts are x = 1 and x = 3. The test point (2, 5)

does not satisfy the inequality.

6. D; The x-intercepts are x = −1 and x = −3. The test point

(−2, 5) satisfi es the inequality.

7. Step 1 Graph y = −x2. Because the inequality symbol is <,

make the parabola dashed.

Step 2 Test a point inside the parabola, such as (0, −1).

y < −x2

−1 <? −02

−1 < 0

So, (0, −1) is a solution of the inequality.


y

42−2−4

−4

−2

2

4

x

8. Step 1 Graph y = 4x2. Because the inequality symbol is ≥,

make the parabola solid.


y ≥ 4x2

1 ≥? 4(0)2

1 ≥ 0



y

2−2

−2

2

4

6

8

x

9. Step 1 Graph y = x2 − 9. Because the inequality symbol

is >, make the parabola dashed.


y > x2 − 9

1 ≥? (0)2 − 9

1 ≥ −9



y

4 62−2−4−6

−4

−6

−10

−2

2

x


Chapter 3

10. Step 1 Graph y = x2 + 5. Because the inequality symbol

is <, make the parabola dashed.


y < x2 + 5

1 <? (0)2 + 5

1 < 5



y

2−2

2

4

6

8

10

12

x

11. Step 1 Graph y = x2 + 5x. Because the inequality symbol

is ≤, make the parabola solid.

Step 2 Test a point outside the parabola, such as (1, −1).

y ≤ x2 + 5x

−1 <? (1)2 + 5(1)

−1 < 6

So, (1, −1) is a solution of the inequality.


y

2−2−4−6

−6

2

4

x

12. Step 1 Graph y = −2x2 + 9x − 4. Because the inequality

symbol is ≥, make the parabola solid.


y ≥ −2x2 + 9x − 4

0 ≥? −2(0)2 + 9(0) − 4

0 ≥ −4



y

62−2

−2

2

4

6

8

x

13. Step 1 Graph y = 2(x + 3)2 − 1. Because the

inequality symbol is >, make the parabola dashed.

Step 2 Test a point inside the parabola, such as (−3, 1).

y > 2(x + 3)2 − 1

1 >? 2(−3 + 3)2 − 1

1 > −1

So, (−3, 1) is a solution of the inequality.


y

−4−6

−2

2

4

6

8

x


Chapter 3

14. Step 1 Graph y = ( x − 1 —

2 ) 2 +

5 —

2 . Because the inequality

symbol is ≤, make the parabola solid.


y ≤ ( x − 1 —

2 )

2

+ 5 —

2

0 ≤? ( 0 −

1 —

2 )

2

+ 5 —

2

0 ≤ 11

— 4



y

−4 −2

2

4

6

8

x2 4

15. Let P = (x1, y1), then the inequality is y1 > f (x1).

16. Let P = (x1, y1), then the inequality is y1 > f (x1).

17. The graph should be solid, not dashed.

y

−4 −2

4

x2 4

18. The region inside the parabola should be shaded, not outside

the parabola.

y

−4 −2

4

x2 4

19. Graph W = 115x2 for nonnegative values of x. Because the

inequality symbol is ≤, make the parabola solid. Test a point

outside the parabola, such as (1, 100).

W ≤ 115x2

100 ≤? 115(1)2

100 ≤ 115

Because (1, 100) is a solution, shade the region outside the

parabola. The shaded region represents weights that can be

supported by shelves with various thicknesses.

2 4 60 x

w

2000

3000

4000

5000

10000

Thickness (in.)

Wei

gh

t (l

bs)

Hardwood Shelf

w ≤ 115x2

20. Graph W = 8000d2 for nonnegative values of d. Because the

inequality symbol is ≤, make the parabola solid, Test a point

outside the parabola, such as (1, 7000).

W ≤ 8000d2

7000 ≤? 8000(1)2

7000 ≤ 8000

Because (1, 7000) is a solution, shade the region outside the

parabola. The shaded region represents weights that can be

supported by wire ropes with various diameters.

2 4 6 8 100 x

w

400,000

600,000

800,000

1,000,000

200,0000

Diameter (in.)

Wei

gh

t (l

bs)

Wire Rope

w ≤ 8000d2

21. Step 1 Graph y ≥ 2x2.

Step 2 Graph y < −x2 + 1.



y

2−2

−2

2

x

y ≥ 2x2

y < −x2 + 1


Chapter 3

22. Step 1 Graph y > −5x2.

Step 2 Graph y > 3x2 − 2.



y

2−2

−6

2

4

x

y > −5x2

y > 3x2 − 2

23. Step 1 Graph y ≤ −x2 + 4x − 4.

Step 2 Graph y < x2 + 2x − 8.



y

4 6 8−2

−4

−2

2

x

y ≤ −x2 + 4x − 4

y < x2 + 2x − 8

24. Step 1 Graph y ≥ x2 − 4.

Step 2 Graph y ≤ −2x2 + 7x + 4.



y

3 6−4

4

6

8

x

10

y ≤ −2x2 + 7x + 4

y ≥ x2 − 4

25. Step 1 Graph y ≥ 2x2 + x − 5.

Step 2 Graph y < −x2 + x + 10.



y

2 4 6−4

4

12

16

x

y ≥ 2x2 + x − 5

y < −x2 + 5x + 10

26. Step 1 Graph y ≥ x2 − 3x − 6.

Step 2 Graph y ≥ x2 + 7x + 6.



y

2 4 6−4−8 −2

6

8

x

−6

−8 y ≥ x2 − 3x − 6y ≥ x2 + 7x + 6


Chapter 3

27. First, write and solve the equation obtained by replacing <

with =.

4x2 = 25

x2 = 25

— 4

x = 5 —

2 or x = − 5 —

2

The numbers − 5 — 2 and

5 —

2 are critical values of the original

inequality. Plot − 5 — 2 and

5 —

2 on a number line, using open dots

because the values do not satisfy the inequality. The critical



inequality.

−


52

0 1 2 3 4 5−1−2−3−4−5

Test x = 3.

52

4(−3)2 ≮ 25

4(0)2 < 25 ✓

4(3)2 ≮ 25

So, the solution is − 5 — 2 < x < 5 —

2 .


< with =.

x2 + 10x + 9 = 0

(x + 9)(x + 1) = 0

x = −9 or x = −1

The numbers −9 and −1 are critical values of the original

inequality. Plot −9 and −1 on a number line, using open





Test x = −5.Test x = −10.

−9 −1

−4 −2 0−6−8−10

Test x = 0.

(−10)2 + 10(−10) + 9 ≮ 0

(−5)2 + 10(−5) + 9 < 0 ✓

02 + 10(0) + 9 ≮ 0

So, the solution is −9 < x < −1.

29. First, write and solve the equation obtained by replacing ≥

with =.

x2 − 11x = −28

x2 − 11x + 28 = 0

(x − 4)(x − 7) = 0

x = 4 or x = 7

The numbers 4 and 7 are critical values of the original

inequality. Plot 4 and 7 on a number line, using closed dots

because the values do satisfy the inequality. The critical



inequality.

6 8 1042 5 7 9310

Test x = 0. Test x = 5. Test x = 8.

02 − 11(0) ≥ −28 ✓

52 − 11(5) ≱ −28

82 − 11(8) ≥ −28 ✓

So, the solution is x ≤ 4 or x ≥ 7.

30. First, write and solve the equation obtained by replacing > with =.

3x2 − 13x = −10

3x2 − 13x + 10 = 0 (3x − 10)(x − 1) = 0

x = 10 —

3 or x = 1

The numbers 10

— 3 and 1 are critical values of the original

inequality. Plot 10

— 3 and 1 on a number line, using open dots




inequality.

Test x = 2.

Test x = 0. 103

0 1 2 3 4 5−1

Test x = 4.

3(0)2 − 13(0) > −10 ✓

3(2)2 − 13(2) ≯ −10

3(4)2 − 13(4) > −10 ✓

So, the solution is x < 1 or x > 10 —

3 .


Chapter 3

31. First, write and solve the equation obtained by replacing ≤ with =.

2x2 − 5x − 3 = 0 (2x + 1)(x − 3) = 0 x = − 1 —

2 or x = 3

The numbers − 1 — 2 and 3 are critical values of the original

inequality. Plot − 1 — 2 and 3 on a number line, using closed

dots because the values do satisfy the inequality. The critical



inequality.

Test x = 0.

Test x = −1. 12

0 1 2 3 4 5 6−1−2−3−4

Test x = 4.−

2(−1)2 − 5(−1) − 3 ≰ 0 2(0)2 − 5(0) − 3 ≤ 0 ✓

2(4)2 − 5(4) − 3 ≰ 0

So, the solution is − 1 — 2 ≤ x ≤ 3.

32. First, write and solve the equation obtained by replacing ≥

with =.

4x2 + 8x − 21 = 0 (2x + 7)(2x − 3) = 0

x = − 7 — 2 or x = 3 —

2

The numbers − 7 — 2 and

3 —

2 are critical values of the original

inequality. Plot − 7 — 2 and

3 —

2 on a number line, using closed dots

because the values do satisfy the inequality. The critical x-values



Test x = 0.

Test x = −4. 72

0 1 2 3 4−1−2−3−4−5−6

Test x = 2.32−

4(−4)2 + 8(−4) − 21 ≥ 0 ✓

4(0)2 + 8(0) − 21 ≱ 0 4(2)2 + 8(2) − 21 ≥ 0 ✓

So, the solution is x ≤ − 7 — 2 or x ≥

3 —

2 .

33. First, write and solve the equation obtained by replacing > with =.

1 —

2 x2 − x = 4

1 —

2 x2 − x − 4 = 0

x2 − 2x − 8 = 0 (x − 4)(x + 2) = 0 x = 4 or x = −2

The numbers −2 and 4 are critical values of the original

inequality. Plot −2 and 4 on a number line, using open dots




inequality.

Test x = 0.

Test x = −3.

0 1 2 3 4 5 6−1−2−3−4

Test x = 5.

1 —

2 (−3)2 − (−3) > 4 ✓

1 —

2 (0)2 − 0 ≯ 4

1 —

2 (5)2 − 5 > 4 ✓

So, the solution is x < −2 or x > 4.

34. First, write and solve the equation obtained by replacing ≤ with =.

− 1 — 2 x2 + 4x = 1

− 1 — 2 x2 + 4x − 1 = 0

x2 − 8x + 2 = 0

x = 8 ± √—

56 —

2

x = 4 ± √—

14

The numbers 4 − √—

14 and 4 + √—

14 are critical values of the

original inequality. Plot 4 − √—

14 and 4 + √—

14 on a number

line, using closed dots because the values do satisfy the

inequality. The critical x-values partition the number line into

three intervals. Test an x-value in each interval to determine

whether it satisfi es the inequality.

Test x = 1.Test x = 0.

420 6 8 10 12−2

Test x = 8.

4 − 14 4 + 14

− 1 — 2 (0)2 + 4(0) ≤ 1 ✓

− 1 — 2 (1)2 + 4(1) ≰ 1

− 1 — 2 (8)2 + 4(8) ≤ 1 ✓

So, the solution is x ≤ 4 − √—

14 or x ≥ 4 + √—

14 .


Chapter 3

35. The solution consists of the x-values for which the graph of

y = x2 − 3x + 1 lies below the x-axis. Find the x-intercepts

of the graph by letting y = 0 and use the Quadratic Formula

to solve 0 = x2 − 3x + 1 for x.

x = −(−3) ± √——

(−3)2 − 4(1)(1) ———

2(1)

x = 3 ± √—

5 —

2

The solutions are x ≈ 0.38 and x ≈ 2.62. Sketch a parabola

that opens up and has 0.38 and 2.62 as x-intercepts. The

graph lies below the x-axis to the right of x = 0.38 and

to the left of x = 2.62. The solution of the inequality is

approximately 0.38 < x < 2.62.

x

y

2

−2

−2


y = x2 − 4x + 2 lies above the x-axis. Find the x-intercepts


to solve 0 = x2 − 4x + 2 for x.

x = −(−4) ± √——

(−4)2 − 4(1)(2) ———

2(1)

x = 4 ± √—

8 —

2

x = 2 ± √—

2

The solutions are x ≈ 0.59 and x ≈ 3.41. Sketch a parabola

that opens up and has 0.59 and 3.41 as x-intercepts. The

graph lies above the x-axis to the left of x = 0.59 and to

the right of x = 3.41. The solution of the inequality is

approximately x < 0.59 or x > 3.41.

x

y

2

−2

2 4

37. The solution consists of the x-values for which the

graph of y = x2 + 8x + 7 lies above the x-axis. Find the

x-intercepts of the graph by letting y = 0 and use factoring to

solve 0 = x2 + 8x + 7 for x.

x2 + 8x + 7 = 0 (x + 7)(x + 1) = 0 x + 7 = 0 or x + 1 = 0 x = −7 or x = −1

The solutions are x = −7 and x = −1. Sketch a parabola

that opens up and has −7 and −1 as x-intercepts. The graph

lies above the x-axis to the left of x = −7 and to the right

of x = −1. The solution of the inequality is

x < −7 or x > −1.

x

y

8

4

−4−8


y = x2 + 6x + 3 lies below the x-axis. Find the x-intercepts


to solve 0 = x2 + 6x + 3 for x.

x = −6 ± √——

62 − 4(1)(3) ——

2(1)

x = −6 ± √—

24 —

2

x = −3 ± √—

6

The solutions are x ≈ −5.45 and x ≈ −0.55. Sketch

a parabola that opens up and has −5.45 and −0.55 as

x-intercepts. The graph lies below the x-axis to the right of

x = −5.45 and to the left of x = −0.55. The solution of the

inequality is approximately −5.45 < x < −0.55.

x

y

4

−4

−2−4−6


Chapter 3


graph of y = 3x2 + 2x − 8 lies on or below the x-axis.

Find the x-intercepts of the graph by letting y = 0 and use

factoring to solve 0 = 3x2 + 2x − 8 for x.

3x2 + 2x − 8 = 0 (3x − 4)(x + 2) = 0 3x − 4 = 0 or x + 2 = 0 x = 4 —

3 or x = −2

The solutions are x = −2 and x = 4 — 3 . Sketch a parabola that

opens up and has −2 and 4 —

3 as x-intercepts. The graph lies on

or below the x-axis to the right of x = −2 and to the left of

x = 4 — 3 . The solution of the inequality is −2 ≤ x ≤ 4 —

3 .

x

y

−4

2


graph of y = 3x2 + 5x − 4 lies below the x-axis. Find

the x-intercepts of the graph by letting y = 0 and use the

Quadratic Formula to solve 0 = 3x2 + 5x − 4 for x.

x = −5 ± √——

52 − 4(3)(−4) ——

2(3)

x = −5 ± √

— 73 —

6

The solutions are x ≈ −2.26 and x ≈ 0.59. Sketch a parabola

that opens up and has −2.26 and 0.59 as x-intercepts. The

graph lies below the x-axis to the right of x = −2.26 and

to the left of x = 0.59. The solution of the inequality is

approximately −2.26 < x < 0.59.

x

y

−4

−2

2−4


graph of y = 1 — 3 x2 + 2x − 2 lies on or above the x-axis. Find

the x-intercepts of the graph by letting y = 0 and use the

Quadratic Formula to solve 0 = 1 — 3 x2 + 2x − 2 for x.

x = −2 ± √

——

22 − 4 ( 1 — 3 ) (−2) ____________________

2 ( 1 — 3 )

x = −2 ± √

—

20

— 3 __________

2 —

3

x = −3 ± √—

15

The solutions are x ≈ −6.87 and x ≈ 0.87. Sketch a parabola

that opens up and has −6.87 and 0.87 as x-intercepts. The

graph lies above the x-axis to the left of x = −6.87 and

to the right of x = 0.87. The solution of the inequality is

approximately x ≤ −6.87 or x ≥ 0.87.

x

y

4

−4

−4−8

42. The solution consists of the x-values for which the graph

of y = 3 — 4 x2 + 4x − 3 lies on or above the x-axis. Find the

x-intercepts of the graph by letting y = 0 and use factoring

to solve 0 = 3 — 4 x2 + 4x − 3 for x.

3 —

4 x2 + 4x − 3 = 0

3x2 + 16x − 12 = 0 (3x − 2) (x + 6) = 0 3x − 2 = 0 or x + 6 = 0 x = 2 —

3 or x = −6

The solutions are x = −6 and x = 2 — 3 . Sketch a parabola that

opens up and has −6 and 2 —

3 as x-intercepts. The graph lies

above the x-axis to the left of x = −6 and to the right of

x = 2 — 3 . The solution of the inequality is x ≤ −6 or x ≥ 2 —

3 .

x

y8

4

−4

4−4−8


Chapter 3

43. a. The solutions of the inequality are x1 < x < x2.

b. The solutions of the inequality are x < x1 or x > x2.

c. If f is refl ected in the x-axis to form g, then

g(x) = −ax2 − bx − c and the graph opens down. The

graph then lies above the x-axis for x1 < x < x2.

44. Let ℓ represent the length (in feet) and w represent the width

(in feet) of the parking lot.

Perimeter = 400 Area ≥ 9100

2ℓ + 2w = 400 ℓw ≥ 9100

Solve the perimeter equation for w to obtain w = 200 − ℓ.

Substitute this into the area inequality to obtain a quadratic

inequality in one variable.

ℓw ≥ 9100

ℓ(200 − ℓ) ≥ 9100

200ℓ − ℓ2 ≥ 9100

−ℓ2 + 200ℓ − 9100 ≥ 0

Use a graphing calculator to fi nd the ℓ-intercept of

y = −ℓ2 + 200ℓ − 9100.

220

−10,000

0

10,000

The ℓ-intercepts are ℓ = 70 and ℓ = 130. The solution

consists of the ℓ-values for which the graph lies on or above

the ℓ-axis. The graph lies on or above the ℓ-axis when

70 ≤ ℓ ≤ 130. So, the length of the parking lot is at least

70 feet and at most 130 feet.

45. To fi nd the distances the arch is above the road, solve

the quadratic inequality 52 ≤ −0.00211x2 + 1.06x. First,

solve the related equation 52 = −0.00211x2 + 1.06x.

52 = −0.00211x2 + 1.06x

0 = −0.00211x2 + 1.06x − 52

x = −1.06 ± √———

1.062 − 4(−0.00211)(−52) ————

2(−0.00211)

x = −1.06 ± √—

0.68472 —— −0.00422

So, the solutions of the equation are approximately

x ≈ 55 and x ≈ 447. Then the solution of the inequality is

approximately 55 < x < 447. So, the arch is above the road

about 55 meters from the left pylon to about 447 meters from

the left pylon.

46. To fi nd when the number of teams is greater than 1000, solve

the quadratic inequality

1000 ≤ 17.155x2 + 193.68x + 235.81. First solve the related

quadratic equation 1000 = 17.155x2 + 193.68x + 235.81.

1000 = 17.155x2 + 193.68x + 235.81

0 = 17.155x2 + 193.68x − 764.19

x = −193.68 ± √———

193.682 − 4(17.155)(−764.19) ————

2(17.155)

x = −193.68 ± √——

89,950.6602 ———

34.31

So, the solutions of the equation are approximately

x ≈ −14.4 and x ≈ 3.1. Reject the negative solution, −14.4,

because time cannot be a negative number. So, after 4 years,

the number of teams was greater than 1000.

47. a. The inequality is A(x) < V(x), or

0.0051x2 − 0.319x + 15 < 0.005x2 − 0.23x + 22

for 16 ≤ x ≤ 70.

b. Graph y1 = 0.0051x2 − 0.319x + 15 and

y2 = 0.005x2 − 0.23x + 22 only on the domain

16 ≤ x ≤ 70.

700

16

40

A(x) is always less than V(x). So, the solution of the

inequality is 16 ≤ x ≤ 70.

c. A driver would react more quickly to the siren of an

approaching ambulance because the graph shows that the

reaction time for audio stimuli is always less than that of

visual stimuli.

48. a. Sample answer: Two solutions are (2, 0) and (3, 1).

b. The points (1, −2) and (5, 6) are not solutions of the

system because the parabolas are dashed, meaning that the

points on the parabola are not included in the solution set.

c. You cannot change the inequality for just one point to be

included. Because both points are points of intersection,

they are either both solutions or both not solutions.


Chapter 3

49. To fi nd the number of days the larvae have a length greater

than 10 millimeters, solve the inequality

10 < 0.00170x2 + 0.145x + 2.35 for 0 ≤ x ≤ 40. First solve

the equation 10 = 0.00170x2 + 0.145x + 2.35.

10 = 0.00170x2 + 0.145x + 2.35

0 = 0.00170x2 + 0.145x − 7.65

x = −0.145 ± √———

0.1452 − 4(0.00170)(−7.65) ————

2(0.00170)

x = −0.145 ± √—

0.073045 ——

0.00340

The solutions of the equation are x ≈ −122 and x ≈ 37.

Reject the negative solution because it falls outside of the

domain of the function. Because the domain is 0 ≤ x ≤ 40,

the ages that the larvae are longer than 10 millimeters are

from 37 to 40 days.

50. Your friend is correct. Because the graphs of the parabolas of

the system both open up, the graphs will intersect in at least

one point. Therefore, the two inequalities have a solution.

51. a. To fi nd the area, you need to fi nd the distance from the

line to the vertex for the height and the distance between

the intersection of the parabola and the line for the base.

First, fi nd the vertex.

x = − b — 2a

= − 4 —

2(−1) = 2

y = −x2 + 4x = −(2)2 + 4(2) = 4 So, the vertex of the parabola is (2, 4). Thus, the height is

∣ 4 − 0 ∣ = 4 units. To fi nd the length of the base, fi nd the

points of intersection of the two related equations.

−x2 + 4x = 0 x(−x + 4) = 0 x = 0 or x = 4 The points of intersection are (0, 0) and (4, 0). Thus, the

length of the base is ∣ 4 − 0 ∣ = 4 units. So, the area is

2 —

3 (4)(4) = 32

— 3 ≈ 10.67 square units.

b. To fi nd the area, you need to fi nd the distance from the

line to the vertex for the height and the distance between

the intersection of the parabola and the line for the base.

First, fi nd the vertex.

x = − b — 2a

= − −4 —

2(1) = 2

y = (2)2 − 4(2) − 5 = −9

So, the vertex of the parabola is (2, −9). Thus, the height

is ∣ −9 − 7 ∣ = 16 units. To fi nd the length of the base, fi nd

the points of intersection of the two related equations.

x2 − 4x − 5 = 7 x2 − 4x − 12 = 0 (x − 6)(x + 2) = 0 x − 6 = 0 or x + 2 = 0 x = 6 or x = −2

The points of intersection are (−2, 7) and (6, 7). Thus, the

length of the base is ∣ −2 − 6 ∣ = 8 units. So, the area is

2 —

3 (16)(8) = 256

— 3 ≈ 85.33 square units.

52. Sample answer:

y

4 6 82−2−4−6−8

2

4

6

8

10

x

y ≤ 112x2 + 3

y ≥ 18x2

The intersection of y ≤ 1 — 12

x2 + 3 and y ≥ 1 — 8 x2 create a shape

similar to a smile, which could be used by a company that

sells toothpaste.

53. a. To determine whether the truck can fi t under the arch,

solve the inequality −0.0625x2 + 1.25x + 5.75 < 11.

First, solve the related quadratic equation

−0.0625x2 + 1.25x + 5.75 = 11.

−0.0625x2 + 1.25x + 5.75 = 11

−0.0625x2 + 1.25x − 5.25 = 0 x2 − 20x + 84 = 0 (x − 14)(x − 6) = 0 x − 14 = 0 or x − 6 = 0 x = 14 or x = 6 The points on the parabola that are exactly 11 feet

high are (6, 11) and (14, 11). These points are

∣ 6 − 14 ∣ = 8 feet apart, so there is enough room for the

7-foot wide truck.


Chapter 3

b. From part(a), the maximum width of an 11-foot truck that

will fi t under the arch is ∣ 14 − 6 ∣ = 8 feet.

c. To fi nd the maximum height of a truck 7 feet wide, fi rst

fi nd the vertex of the parabola, then fi nd the points on the

parabola 3.5 feet from the axis of symmetry to determine

the height.

x = − b — 2a

= − 1.25 —

2(−0.0625) = 10

y = −0.0625(10)2 + 1.25(10) + 5.75 = 12

The vertex of the parabola is (10, 12). Next, fi nd the

height of the parabola 3.5 feet from the axis of symmetry.

So, compute y when x = 10 + 3.5 = 13.5.

y = −0.0625(13.5)2 + 1.25(13.5) + 5.75 ≈ 11.2

So, the maximum height of a truck that is 7 feet wide that

will fi t under the arch is about 11.2 feet.


54. The function is in intercept form, so the x-intercepts are

(−7, 0) and (9, 0). To fi nd the y-intercept, let x = 0 and solve

for y.

y = (0 + 7) (0 − 9) = 7(−9) = −63

So, the y-intercept is (0, −63).

y

4 6 82−2−4−6

−80

−40

−20

x

(−7, 0) (9, 0)

(0, −63)

55. To fi nd the x-intercept(s), let y = 0 and solve for x.

0 = (x − 2)2 −4

4 = (x − 2)2

±2 = x − 2 2 ± 2 = x x = 0 or x = 4

So, the x-intercepts are (0, 0) and (4, 0). The y-intercept is

also (0, 0).

y

62−2

−4

−2

4

6

x

(0, 0) (4, 0)

56. To fi nd the x-intercept(s), let y = 0 and solve for x.

0 = −x2 + 5x − 6 0 = x2 − 5x + 6 0 = (x − 2)(x − 3)

x − 2 = 0 or x − 3 = 0 x = 2 or x = 3 So, the x-intercepts are (2, 0) and (3, 0). To fi nd the

y-intercept, let x = 0 and solve for y.

y = 02 + 5(0) − 6 = −6

So, the y-intercept is (0, −6).

y

62 4−2

−4

−6

−2

x

(2, 0)

(0, −6)

(3, 0)

57. Because a < 0, the parabola opens down and the

function has a maximum value. Find the vertex of the

parabola. First, fi nd the x-coordinate.

x = − b — 2a

= − −6 —

2(−1) = −3

Next, fi nd the y-coordinate of the vertex.

f (−3) = −(−3)2 − 6(−3) − 10 = −1

The vertex of the parabola is (−3, −1). So, the maximum

value of the function is −1. The function is increasing to the

left of x = −3 and decreasing to the right of x = −3.

58. Because a > 0, the parabola opens up and the function has

a minimum value. Find the vertex of the parabola. Because

the function is in vertex form, the vertex is (−2, −1). So,

the minimum value of the function is −1. The function is

decreasing to the left of x = −2 and increasing to the right

of x = −2.

59. Because a < 0, the parabola opens down and the function

has a maximum value. Find the vertex of the parabola. The

x-intercepts of the function are 3 and −7. Find the vertex of

the function. First, fi nd the x-coordinate.

x = p + q —

2 = − 3 + (−7)

— 2 = −2


f (−2) = −(−2 − 3)(−2 + 7) = 25

The vertex of the parabola is (−2, 25). So, the maximum

value of the function is 25. The function is increasing to the

left of x = −2 and decreasing to the right of x = −2.


Chapter 3

60. Because a > 0, the parabola opens up and the function has

a minimum value. Find the vertex of the parabola. First, fi nd

the x-coordinate.

x = − b — 2a

= − 3 —

2(1) = − 3 —

2


h ( − 3 — 2 ) = ( − 3 —

2 ) 2 + 3 ( − 3 —

2 ) − 18 = − 81

— 4

The vertex of the parabola is ( − 3 — 2 , − 81

— 4 ) . So, the minimum

value of the function is − 81 —

4 . The function is decreasing

to the left of x = − 3 — 2 and increasing to the right of x = − 3 —

2 .

3.4–3.6 What Did You Learn? (p. 147)

1. You can use a graphing calculator to graph both functions.

The graph that has a positive x-intercept with a lesser value

is the one that lands fi rst.

2. Sample answer: A question to ask is: How many points of

intersection are possible when a line intersects a circle?

3. Set up an equation for the perimeter of the fountain

2ℓ + 2w = 400. Write an inequality for the area

ℓw > 9100. Solve the equation for the perimeter for one

variable, and then substitute the result into the inequality for

the area. Solve the resulting inequality.

Chapter 3 Review (pp. 148–150)


y = x2 − 2x − 8.

x

y4

−8

−4


x = −2 and x = 4.

2. 3x2 − 4 = 8 3x2 = 12

x2 = 4 x = ± √

— 4

x = ±2

So, the solutions of the equation are x = −2 and x = 2.

3. x2 + 6x − 16 = 0 (x + 8)(x − 2) = 0

x + 8 = 0 or x − 2 = 0 x = −8 or x = 2 So, the solutions of the equation are x = −8 and x = 2.

4. 2x2 − 17x = −30

2x2 − 17x + 30 = 0 (2x − 5)(x − 6) = 0 2x − 5 = 0 or x − 6 = 0

x = 5 — 2 or x = 6

So, the solutions of the equation are x = 5 — 2 and x = 6.

5. An equation that represents the area is

2 ⋅ 35 ⋅ 18 = (x + 35)(x + 18) because the same amount

will be added to both the width and length and the area is

doubled. Solve the equation.

2 ⋅ 35 ⋅ 18 = (x + 35)(x + 18)

1260 = x2 + 53x + 630

0 = x2 + 53x − 630

0 = (x − 10)(x + 63)

x − 10 = 0 or x + 63 = 0 x = 10 or x = −63

Reject the negative solution because measurements are

positive. Use x = 10. So, the dimensions of the new area are

45 feet by 28 feet.



36 = 4x −y = 3 x = 9 y = −3

So, x = 9 and y = −3.

7. (−2 + 3i) + (7 − 6i) = (−2 + 7) + (3 − 6)i

= 5 − 3i

8. (9 + 3i) − (−2 − 7i) = (9 + 2) + (3 + 7)i

= 11 + 10i

9. (5 + 6i)(−4 + 7i) = −20 + 35i − 24i + 42i2

= −20 + 11i + 42(−1)

= −62 + 11i

10. 7x2 + 21 = 0 7x2 = −21

x2 = −3

x = ± √—

−3

x = ±i √—

3

The solutions are x = i √—

3 and x = −i √—

3 .


Chapter 3

11. 2x2 + 32 = 0 2x2 = −32

x2 = −16

x = ± √—

−16

x = ±4i

So, the zeros of f are 4i and −4i.


y = −16t2 + 96t + 4 y = −16(t2 − 6t) + 4 y + ? = −16(t2 − 6t + ?) + 4 y + (−16)(9) = −16(t2 − 6t + 9) + 4 y − 144 = −16(t − 3)2 + 4 y = −16(t − 3)2 + 148

The vertex is (3, 148). So, the maximum height of the T-shirt

is 148 feet.

13. x2 + 16x + 17 = 0 x2 + 16x = −17

x2 + 16x + 64 = −17 + 64

(x + 8)2 = 47

x + 8 = ± √—

47

x = −8 ± √—

47


47 and x = −8 + √—

47 .

14. 4x2 + 16x + 25 = 0

x2 + 4x + 25 —

4 = 0

x2 + 4x = − 25 —

4

x2 + 4x + 4 = − 25 —

4 + 4

(x + 2)2 = − 9 — 4

x + 2 = ± √—

− 9 — 4

x = −4 ± 3i —

2

The solutions are x = −4 − 3i —

2 and x = −4 + 3i

— 2 .

15. 9x(x − 6) = 81

x(x − 6) = 9 x2 − 6x = 9 x2 − 6x + 9 = 9 + 9 (x − 3)2 = 18

x − 3 = ± √—

18

x = 3 ± 3 √—

2

The solutions are x = 3 −3 √—

2 and x = 3 + 3 √—

2 .

16. y = x2 − 2x + 20

y + ? = (x2 − 2x + ?) + 20

y + 1 = (x2 − 2x + 1) + 20

y + 1 = (x − 1)2 + 20

y = (x − 1)2 + 19

The vertex form of the function is y = (x − 1)2 + 19. The

vertex is (1, 19).

17. −x2 + 5x = 2 −x2 + 5x − 2 = 0

x = −b ± √—

b2 − 4ac ——

2a

x = −5 ± √——

52 − 4(−1)(−2) ———

2(−1)

x = 5 ± √—

17 —

2

So, the solutions are x = 5 − √—

17 —

2 and x = 5 + √

— 17 —

2 .

18. 2x2 + 5x = 3 2x2 + 5x − 3 = 0

x = −b ± √

— b2 − 4ac ——

2a

x = −5 ± √——

52 − 4(2)(−3) ——

2(2)

x = −5 ± √—

49 —

4

x = −5 ± 7 —

4

So, the solutions are x = 1 — 2 and x = −3.

19. 3x2 − 12x + 13 = 0

x = −b ± √—

b2 − 4ac ——

2a

x = −(−12) ± √——

(−12)2 − 4(3)(13) ———

2(3)

x = 12 ± √—

−12 —

6

x = 12 ± 2i √—

3 —

6

x = 6 ± i √—

3 —

3

So, the solutions are x = 6 + i √—

3 —

3 and x = 6 − i √

— 3 —

3 .

20. Equation: −x2 − 6x − 9 = 0 Discriminant: b2 − 4ac = (−6)2 − 4(−1)(−9) = 0 The equation has one real soluton.


— b2 − 4ac ——

2a =

−(−6) ± √—

0 ——

2(−1) = −3


Chapter 3

21. Equation: x2 − 2x − 9 = 0 Discriminant: b2 − 4ac = (−2)2 − 4(1)(−9) = 40



— b2 − 4ac ——

2a

x = −(−2) ± √

— 40 ——

2(1)

x = 1 ± √—

10

22. Equation: x2 + 6x + 5 = 0 Discriminant: b2 − 4ac = 62 − 4(1)(5) = 16



— b2 − 4ac ——

2a

x = −6 ± √

— 16 —

2(1)

x = −3 ± 2

x = −5 or x = −1

23. Use substitution because both equations are already solved

for y. Substitute −2x + 2 for y in Equation 1 and solve for x.

2x2 − 2 = −2x + 2 2x2 + 2x − 4 = 0 x2 + x − 2 = 0 (x + 2)(x − 1) = 0 x + 2 = 0 or x − 1 = 0 x = −2 or x = 1 To solve for y, substitute x = −2 and x = 1 into the equation

y = −2x + 2.

y = −2x + 2 = −2(−2) + 2 = 6 y = −2x + 2 = −2(1) + 2 = 0 The solutions are (−2, 6) and (1, 0).

24. Use elimination because adding like terms will result in a

quadratic equation in one variable. First, add the equations to

eliminate the y term and obtain a quadratic equation in x.

y = x2 − 6x + 13

−y = − 2x + 3

0 = x2 − 8x + 16

Solve by factoring.

x2 − 8x + 16 = 0 (x − 4)2 = 0 x − 4 = 0 x = 4 To solve for y, substitute x = 4 into the equation y = 2x − 3.

y = 2x − 3 = 2(4) − 3 = 5 The solution is (4, 5).

25. Use substitution because elimination is not a possibility with

no like terms. First, solve for y in Equation 2.

y = −3x + 1 Next, substitute −3x + 1 for y in Equation 1 and solve for x.

x2 + (−3x + 1)2 = 4

x2 + 9x2 − 6x + 1 = 4

10x2 − 6x − 3 = 0

x = −(−6) ± √——

(−6)2 − 4(10)(−3) ———

2(10)

x = 6 ± √—

156 —

20

x = 6 ± 2 √—

39 —

20

x = 3 ± √—

39 —

10

To solve for y, substitute x = 3 − √—

39 —

10 and x = 3 + √

— 39 —

10

into the equation y = −3x + 1.

y = −3x + 1 = −3 ( 3 − √—

39 —

10 ) + 1

= −9 + 3 √—

39 —

10 + 1 = 1 + 3 √

— 39 —

10

y = −3x + 1 = −3 ( 3 + √—

39 —

10 ) + 1

= −9 − 3 √—

39 —

10 + 1 = 1 − 3 √

— 39 —

10

The solutions are ( 3 − √—

39 —

10 ,

1 + 3 √—

39 —

10 ) and

( 3 + √—

39 —

10 ,

1 − 3 √—

39 —

10 ) .

26. Write a system of equations using each side of the

original equation.

Equation System

−3x2 + 5x − 1 = 5x2 − 8x − 3 y = −3x2 + 5x − 1 y = 5x2 − 8x − 3 Graph the equations in the same plane.

x

y

−2

4 6−2

(−0.14, −1.77)

(1.77, −1.53)

The solutions are x ≈ −0.14 and x ≈ 1.77.


Chapter 3

27. Step 1 Graph y = x2 + 8x + 16. Because the inequality

symbol is >, make the parabola dashed.


y > x2 + 8x + 16

1 >? (−4)2 + 8(−4) + 16

1 > 0 So, (−4, 1) is a solution of the inequality.


y

2−2−4−6−8−10

−2

2

4

6

8

x

y > x2 + 8x + 16

28. Step 1 Graph y = x2 + 6x + 8. Because the inequality

symbol is ≥, make the parabola with a solid line.


y ≥ x2 + 6x + 8 1 ≥

? (−3)2 + 6(−3) + 8

1 ≥ −1

So, (−3, 1) is a solution of the inequality.


2−4−6−8

−2

2

4

8

x

y

y ≥ x2 + 6x + 8

29. Step 1 Rewrite the inequality.

x2 + y ≤ 7x − 12

y ≤ −x2 + 7x − 12

Step 2 Graph y = −x2 + 7x − 12. Because the inequality

symbol is ≤, make the parabola with a solid line.

Step 3 Test a point inside the parabola, such as (3, −1).

y ≤ −x2 + 7x − 12

−1 ≤? −(3)2 + 7(3) − 12

−1 ≤ 0 So, (3, −1) is a solution of the inequality.


y

4 6 82−2

−4

−6

−8

−2

2

x

y ≤ −x2 + 7x − 12

30. Step 1 Graph x2 − 4x + 8 > y.

Step 2 Graph −x2 + 4x + 2 ≤ y.



x

y

6

4

2

8

42−2 6

y < x2 − 4x + 8

y ≥ −x2 + 4x + 2


Chapter 3

31. Step 1 Rewrite the system.

2x2 − x + 5 ≥ y 0.5x2 + 2x − 1 > y Step 2 Graph 2x2 − x + 5 ≥ y.

Step 3 Graph 0.5x2 + 2x − 1 > y.



x

y

4

10

8

2

62 4−2−4−6−8

y < 0.5x2 + 2x + 1

y ≤ 2x2 − x + 5

32. Step 1 Rewrite the system.

−3x2 − 2x − 1 ≤ y 2x2 − x + 5 < y Step 2 Graph −3x2 − 2x − 1 ≤ y.

Step 3 Graph 2x2 − x + 5 < y.



y

4 62−2−4−6

−4

−6

−8

2

4

8

x

y > 2x2 − x + 5

y ≥ −3x2 − 2x − 1

33. First, write and solve the equation obtained by

replacing ≥ with = . 3x2 + 3x − 60 = 0 x2 + x − 20 = 0 (x + 5)(x − 4) = 0 x = −5 or x = 4 The numbers −5 and 4 are critical values of the original

inequality. Plot −5 and 4 on a number line, using closed dots

because the values do satisfy the inequality. The critical x-values



Test x = −6.

−5

0 2 4 6−2−4−6−8

Test x = 5.Test x = 0.

3(−6)2 + 3(−6) − 60 ≥ 0 ✓

3(0)2 + 3(0) − 60 ≱ 0 3(5)2 + 3(5) − 60 ≥ 0 ✓

So, the solution is x ≤ −5 or x ≥ 4.

34. First, write and solve the equation obtained by

replacing < with = . −x2 − 10x = 21

x2 + 10x + 21 = 0 (x + 3)(x + 7) = 0 x = −3 or x = −7

The numbers −7 and −3 are critical values of the original

inequality. Plot −7 and −3 on a number line, using open





Test x = −4.

−3 −2 −1 0−4−5−6−7−8−9−10


−(−8)2 − 10(−8) < 21 ✓

−(−4)2 − 10(−4) ≮ 21

−(0)2 − 10(0) < 21 ✓

So, the solution is x < −7 or x > −3.


Chapter 3


≤ with = . 3x2 + 2 = 5x

3x2 − 5x + 2 = 0 (3x − 2)(x − 1) = 0 3x − 2 = 0 or x − 1 = 0 x = 2 —

3 or x = 1

The numbers 2 —

3 and 1 are critical values of the original

inequality. Plot 2 —

3 and 1 on a number line, using closed dots

because the values do satisfy the inequality. The critical



inequality.

0 113

Test x = .Test x = 0.

23

43

43Test x = .3

4

3(0)2 + 2 ≰ 5(0)

3 ( 3 — 4 )

2

+ 2 ≤ 5 ( 3 — 4 ) ✓

3 ( 4 — 3 )

2

+ 2 ≰ 5 ( 4 — 3 )

So, the solution is 2 —

3 ≤ x ≤ 1.

Chapter 3 Test (p.151)

1. Use the process of completing the square because a = 1 and

b is an even number.

0 = x2 + 2x + 3 −3 = x2 + 2x

−3 + 1 = x2 + 2x + 1 −2 = (x + 1)2

±i √—

2 = x + 1 −1 ± i √

— 2 = x

The solutions are x = −1 − i √—

2 and x = −1 + i √—

2 .

2. Use the process of completing the square because a = 1 and

b is an even number.

6x = x2 + 7 −7 = x2 − 6x

−7 + 9 = x2 − 6x + 9 2 = (x − 3)2

± √—

2 = x − 3 3 ± √

— 2 = x


2 and x = 3 + √—

2 .



x2 + 49 = 85

x2 = 36

x = ±6




(x + 4)( x − 1) = −x2 + 3x + 4 x2 + 3x − 4 = −x2 + 3x + 4 2x2 = 8 x2 = 4 x = ±2


5. The related quadratic equation will have one real solution

because the graph has only one x-intercept. The discriminant

is 32 − 4 ( 1 — 2 ) ( 9 —

2 ) = 0, which indicates one real solution.

6. The related quadratic equation will have two imaginary

solutions because the graph has no x-intercept. The

discriminant is 162 − 4(4)(18) = −32, which indicates two

imaginary solutions.

7. The related quadratic equation will have two real solutions

because the graph has two x-intercepts. The discriminant

is ( 1 — 2 )

2

− 4(−1) ( 3 — 2 ) = 25

— 4 , which indicates two real solutions.


y = 2x − 18

Next, substitute 2x − 18 for y in Equation 1 and solve for x.

x2 + 66 = 16x − (2x − 18)

x2 + 66 = 16x − 2x + 18

x2 + 66 = 14x + 18

x2 − 14x + 48 = 0 (x − 6)(x − 8) = 0 x = 6 or x = 8 To solve for y, substitute x = 6 and x = 8 into the equation

y = 2x − 18.

y = 2x − 18 = 2(6) − 18 = −6

y = 2x − 18 = 2(8) − 18 = −2

The solutions are (6, −6) and (8, −2).


Chapter 3

9. Step 1 Graph y ≥ 1 —

4 x2 − 2.

Step 2 Graph y < −(x + 3)2 + 4.

Step 3 Identify where the two graphs overlap. This region is

the graph of the system.

y

4 6 82−4−6

−4

−6

−8

−10

2

4

x

y < −(x + 3)2 + 4

y ≥ 14x2 − 2


0 = x2 + (x + 4)2 − 40

0 = x2 + x2 + 8x + 16 − 40

0 = 2x2 + 8x − 24

0 = x2 + 4x − 12

0 = (x + 6) (x − 2)

x + 6 = 0 or x − 2 = 0 x = −6 or x = 2 To solve for y, substitute x = −6 and x = 2 into the equation

y = x + 4.

y = x + 4 = −6 + 4 = −2

y = x + 4 = 2 + 4 = 6 The solutions are (−6, −2) and (2, 6).

11. (3 + 4i)(4 − 6i) = 12 − 18i + 16i − 24i2

= 12 − 2i − 24(−1)

= 36 − 2i

12. Using the Pythagorean Theorem, the equation that models

the situation is (16x)2 + (9x)2 = 322. Solve the equation

for x.

(16x)2 + (9x)2 = 322

256x2 + 81x2 = 1024

337x2 = 1024

x2 = 1024 —

337

x = ± √—

1024

— 337

Reject the negative solution, −1.743, because length cannot

be negative. So, x ≈ 1.743. Then the dimensions of the TV

are 16(1.743) ≈ 27.9 inches by 9(1.743) ≈ 15.7 inches.

13. To fi nd what distance the arch is more than 200 feet

above the ground, use the quadratic inequality

200 ≤ −0.0063x2 + 4x. First, rewrite using the related

equation 200 = −0.0063x2 + 4x. Solve the equation.

200 = −0.0063x2 + 4x

0 = −0.0063x2 + 4x − 200

x = −4± √—

10.96 —— −0.0126

So, the solutions are approximately 55 and 580. The arch is

more than 200 feet above the ground from about 55 feet to

about 580 feet.

14. To fi nd the maximum height, fi nd the vertex of the parabola.

Find the x-coordinate fi rst.

x = − b —

2a = −

0.3 —

2(−0.01) = 15


y = −0.01(15)2 + 0.3(15) + 2 = 4.25

The vertex is (15, 4.25). So, the maximum height of the

horseshoe is 4.25 feet. To fi nd how far the horseshoe

traveled, fi nd the x-intercept of the graph.

−0.01x2 + 0.3x + 2 = 0 x2 − 30x − 200 = 0 x = −(−30) ± √

—— (−30)2 − 4(1)(−200) ———

2(1)

x = 30 ± √—

1700 ——

2

x ≈ 35.6 or x ≈ −5.6

Reject the negative solution, −5.6, because distance cannot

be negative. So, the distance the horseshoe traveled is

approximately 35.6 feet.

Chapter 3 Standards Assessment (pp.152–153)

1. B; Factor the right side to obtain (x + 3)(x − 2). So, the

x-intercepts are −3 and 2. Also, the parabola is solid and the

inside of the parabola is shaded.

2. a. The parent function is f (x) = x. The graph of g is a

translation 4 units up of the parent linear function.

b. The parent function is f (x) = 1. The graph of h is a

translation 5 units up of the parent constant function.

c. The parent function is f (x) = x2. The graph of h is a

translation 7 units down of the parent quadratic function.

d. The parent function is f (x) = ∣ x ∣ . The graph of g is a

refl ection in the x-axis, followed by a translation 3 units

left and 9 units down of the parent absolute value

function.

e. The parent function is f (x) = x2. The graph of g is a

vertical shrink by a factor of 1 —

4 followed by a translation

2 units right and 1 unit down of the parent quadratic

function.

f. The parent function is f (x) = x. The graph of h is a

vertical stretch by a factor of 6 followed by a translation

11 units up of the parent linear function.


Chapter 3

3. Because the areas are shaded below the parabolas and

the parabolas are dashed, use the symbol < . y < −0.002x2 + 0.82x + 3.1

y < −0.003x2 + 1.21x + 3.3

4. Your claim is f (x) = −2x2 + 8x + 8 and your friend’s

claim is f (x) = 3x2 + 12x + 8. Your claim will give an axis

of symmetry of x = − b — 2a

= − 8 —

2(−2) = 2 and your friend’s

claim will give an axis of symmetry of

x = − b — 2a

= − 12 —

2(3) = −2.

5. Solve the system by the elimination method. Subtract

the two equations.

y = x2 − 6x + 14

−( y = 2x + 7)

0 = x2 − 8x + 7 0 = (x − 1)(x − 7)

x = 1 or x = 7 So, the x-coordinates of the solutions of the system are

1 and 7.

6. B; Create a scatter plot of the data. The data show a linear

relationship. Sketch the line that must closely appears to fi t

the data. Choose two points on the line, such as (0, 450) and

(10, 350).Write an equation of the line. First, fi nd the slope.

m = 350 − 450 —

10 − 0 = −100 —

10 = −10

Use point-slope form to write an equation. Use

(t1, y1) = (0, 450).

Time (seconds)

Alt

itu

de

(fee

t)

t

y

200

300

400

100

020 30100

y − y1 = m(t − t1) y − 450 = −10(t − 0)

y − 450 = −10t

y = −10t + 450

Use the equation to estimate how long it will take to descend

to an altitude of 100 feet.

100 = −10t + 450

−350 = −10t

35 = t So, it will take 35 seconds to descend to an altitude of

100 feet.

7. x2 + 16 = 0 x2 = −16

x = ± √—

16

x = ±4i

The solutions are x = ±4i. Using the solutions, write the

expression as the product of two binomials:

x2 + 16 = (x − 4i)(x + 4i).

8. Sample answer:

a. x = 1 — 4 (x + 2)2 + 1

b. x = 1 — 4 (x − 2)2 + 1

c. x = 1 — 4 (x + 2)2 −10

9. a. y = x2 − x + 2 y − 2 = x2 − x

y − 2 + 1 — 4 = x2 − x + 1 —

4

y − 7 — 4 = ( x − 1 —

2 )

2

y = ( x − 1 — 2 )

2

+ 7 — 4

This is not a perfect square trinomial.

b. y = x2 + 4x + 6 y − 6 = x2 + 4x

y − 6 + 4 = x2 + 4x + 4 y − 2 = (x + 2)2

y = (x + 2)2 + 2 This is not a perfect square trinomial.

c. y = x2 + 3x + 9 — 4

y − 9 — 4 = x2 + 3x

y − 9 — 4 + 9 —

4 = x2 + 3x + 9 —

4

y = ( x + 3 — 2 )

2

This is a perfect square trinomial.

d. y = x2 − 6x + 9 y − 9 = x2 − 6x

y − 9 + 9 = x2 − 6x + 9 y = (x − 3)2

This is a perfect square trinomial.

Chapter 3 - WHS SALISBURY - Homewhssalisbury.weebly.com/.../8/112801805/hscc_alg2_wsk_03.pdf−8 =...

Documents

Transcript of Chapter 3 - WHS SALISBURY - Homewhssalisbury.weebly.com/.../8/112801805/hscc_alg2_wsk_03.pdf−8 =...