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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-
height ratio of 3:2 and shows the graph touching the x-axis.
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-
height ratio of 3:2 and shows the graph touching the x-axis.
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-
height ratio of 3:2 and shows the graph touching the x-axis.
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.
2. The equation is in standard form. Graph the related function
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.
Copyright © Big Ideas Learning, LLC Algebra 2 91All rights reserved. Worked-Out Solutions
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.
10. To fi nd the zeros of the function, fi nd the x-values for which
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.
92 Algebra 2 Copyright © Big Ideas Learning, LLC Worked-Out Solutions All rights reserved.
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
3. The equation is in standard form. Graph the related function
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.
4. The equation is in standard form. Graph the related function
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.
5. The equation is in standard form. Graph the related function
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
The x-intercepts are −2 and 2. The solutions, or roots, are
x = −2 and x = 2.
7. Rewrite the equation in standard form.
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.
8. Rewrite the equation in standard form.
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.
Copyright © Big Ideas Learning, LLC Algebra 2 93All rights reserved. Worked-Out Solutions
Chapter 3
9. Rewrite the equation in standard form.
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.
10. Rewrite the equation in standard form.
2x = x2 + 2 0 = x2 − 2x + 2 Graph the related function y = x2 − 2x + 2.
x
y
2
2−2
There is no x-intercept. The equation has no real solution.
11. Rewrite the equation in standard form.
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
There is no x-intercept. The equation has no real solution.
12. Rewrite the equation in standard form.
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 .
94 Algebra 2 Copyright © Big Ideas Learning, LLC Worked-Out Solutions All rights reserved.
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.
Copyright © Big Ideas Learning, LLC Algebra 2 95All rights reserved. Worked-Out Solutions
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.
96 Algebra 2 Copyright © Big Ideas Learning, LLC Worked-Out Solutions All rights reserved.
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 .
41. The equation can be written in the form u2 = d, so solve by
using square roots.
−(x + 9)2 = 64
(x + 9)2 = −64
There is no real solution.
42. The equation can be written in the form u2 = d, so solve by
using square roots.
−2(x + 2)2 = 5
(x + 2)2 = − 5 — 2
There is no real solution.
43. The equation can be written in the form u2 = d, so solve by
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.
44. The equation can be factored, so solve by factoring.
t2 + 8t + 16 = 0 (t + 4)2 = 0 t + 4 = 0 t = −4
The solution of the equation is t = −4.
45. The equation can be factored, so solve by factoring.
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 .
46. The equation can be written in the form u2 = d, so solve by
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.
47. To fi nd the zeros of the function, fi nd the x-values for which
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.
48. To fi nd the zeros of the function, fi nd the x-values for which
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.
50. To fi nd the zeros of the function, fi nd the x-values for which
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.
51. To fi nd the zeros of the function, fi nd the x-values for which
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.
Copyright © Big Ideas Learning, LLC Algebra 2 97All rights reserved. Worked-Out Solutions
Chapter 3
52. To fi nd the zeros of the function, fi nd the x-values for which
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 .
53. To fi nd the zeros of the function, fi nd the x-values for which
g(x) = 0.
x2 + 22x + 121 = 0 (x + 11)2 = 0 x + 11 = 0 x = −11
The zero of the function is x = −11.
54. To fi nd the zeros of the function, fi nd the x-values for which
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.
98 Algebra 2 Copyright © Big Ideas Learning, LLC Worked-Out Solutions All rights reserved.
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.
Copyright © Big Ideas Learning, LLC Algebra 2 99All rights reserved. Worked-Out Solutions
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.
100 Algebra 2 Copyright © Big Ideas Learning, LLC Worked-Out Solutions All rights reserved.
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
Copyright © Big Ideas Learning, LLC Algebra 2 101All rights reserved. Worked-Out Solutions
Chapter 3
3.2 Explorations (p. 103)
1. a. √—
9 is in the subsets of real numbers, rational numbers,
integers, whole numbers, and natural numbers.
b. √—
0 is in the subsets of real numbers, rational numbers,
integers, and whole numbers.
c. − √—
4 is in the subsets of real numbers, rational numbers,
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.
3.2 Monitoring Progress (pp. 104–107)
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.
6. Set the real parts equal to each other and the imaginary parts
equal to each other.
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
102 Algebra 2 Copyright © Big Ideas Learning, LLC Worked-Out Solutions All rights reserved.
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
The solutions are −i √—
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
The solutions are −2i √—
7 and 2i √—
7 .
18. 3x2 − 7 = −31
3x2 = −24
x2 = −8
x = ± √—
−8
x = ±2i √—
2
The solutions are −2i √—
2 and 2i √—
2 .
19. 5x2 + 33 = 3 5x2 = −30
x2 = −6
x = ± √—
−6
x = ±i √—
6
The solutions are −i √—
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.
3.2 Exercises (pp. 108–110)
Vocabulary and Core Concept Check
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.
Monitoring Progress and Modeling with Mathematics
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
13. Set the real parts equal to each other and the imaginary parts
equal to each other.
4x = 8 2 = y x = 2 y = 2 So, x = 2 and y = 2.
Copyright © Big Ideas Learning, LLC Algebra 2 103All rights reserved. Worked-Out Solutions
Chapter 3
14. Set the real parts equal to each other and the imaginary parts
equal to each other.
3x = 27 6 = y x = 9 y = 6 So, x = 9 and y = 6.
15. Set the real parts equal to each other and the imaginary parts
equal to each other.
−10x = 20 12 = 3y
x = −2 y = 4 So, x = −2 and y = 4.
16. Set the real parts equal to each other and the imaginary parts
equal to each other.
9x = −36 −18 = 6y
x = −4 y = −3
So, x = −4 and y = −3.
17. Set the real parts equal to each other and the imaginary parts
equal to each other.
2x = 14 −y = 12
x = 7 y = −12
So, x = 7 and y = −12.
18. Set the real parts equal to each other and the imaginary parts
equal to each other.
−12x = 60 y = −13
x = −5
So, x = −5 and y = −13.
19. Set the real parts equal to each other and the imaginary parts
equal to each other.
54 = 9x − 1 — 7 y = −4
x = 6 y = 28
So, x = 6 and y = 28.
20. Set the real parts equal to each other and the imaginary parts
equal to each other.
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.
34. The resistor has a resistance of 4 ohms, so its impedance
is 4 ohms. The inductor has a reactance of 6 ohms, so its
impedance is 6i ohms. The capacitor has a reactance of
9 ohms, so its impedance is −9i ohms.
Impedance of circuit = 4 + 6i + (−9i) = 4 − 3i
The impedance of the circuit is (4 − 3i) ohms.
35. The resistor has a resistance of 8 ohms, so its impedance
is 8 ohms. The inductor has a reactance of 3 ohms, so its
impedance is 3i ohms. The capacitor has a reactance of
2 ohms, so its impedance is −2i ohms.
Impedance of circuit = 8 + 3i + (−2i) = 8 + i The impedance of the circuit is (8 + i) ohms.
104 Algebra 2 Copyright © Big Ideas Learning, LLC Worked-Out Solutions All rights reserved.
Chapter 3
36. The resistor has a resistance of 14 ohms, so its impedance
is 14 ohms. The inductor has a reactance of 7 ohms, so its
impedance is 7i ohms. The capacitor has a reactance of
8 ohms, so its impedance is −8i ohms.
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 .
Copyright © Big Ideas Learning, LLC Algebra 2 105All rights reserved. Worked-Out Solutions
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
106 Algebra 2 Copyright © Big Ideas Learning, LLC Worked-Out Solutions All rights reserved.
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
Maintaining Mathematical Profi ciency
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.
Copyright © Big Ideas Learning, LLC Algebra 2 107All rights reserved. Worked-Out Solutions
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.
83. Use the vertex to fi nd a in the vertex form.
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.
84. Use the vertex to fi nd a in the vertex form.
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.
6. Step 1 Find half the coeffi cient of x. −9
— 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
.
108 Algebra 2 Copyright © Big Ideas Learning, LLC Worked-Out Solutions All rights reserved.
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
The solutions are x = −4 − √—
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
The solutions are x = −3 − √—
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
The solutions are x = 1 − √—
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.
Copyright © Big Ideas Learning, LLC Algebra 2 109All rights reserved. Worked-Out Solutions
Chapter 3
3.3 Exercises (pp. 116–118)
Vocabulary and Core Concept Check
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.
Monitoring Progress and Modeling with Mathematics
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
The solutions are x = 9 − √—
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
Step 3 Replace c with the result of Step 2. x2 + 10x + 25
The expression x2 + 10x + c is a perfect square trinomial
when c = 25. Then
x2 + 10x + 25 = (x + 5)(x + 5) = (x + 5)2.
12. Step 1 Find half the coeffi cient of x. 20
— 2 = 10
Step 2 Square the result of Step 1. (10)2 = 100
Step 3 Replace c with the result of Step 2. x2 + 20x + 100
The expression x2 + 20x + c is a perfect square trinomial
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 2 Square the result of Step 1. (−11)2 = 121
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 .
15. Step 1 Find half the coeffi cient of x. −6
— 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.
16. Step 1 Find half the coeffi cient of x. 24
— 2 = 12
Step 2 Square the result of Step 1. (12)2 = 144
Step 3 Replace c with the result of Step 2. x2 + 24x + 144
The expression x2 + 24x + c is a perfect square trinomial
when c = 144. Then
x2 + 24x + 144 = (x + 12)(x + 12) = (x + 12)2.
110 Algebra 2 Copyright © Big Ideas Learning, LLC Worked-Out Solutions All rights reserved.
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
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
.
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 2 Square the result of Step 1. (−13)2 = 169
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.
22. The value of c is 8 ⋅ 8 = 64; x2 + 16x + 64 = (x + 8)2.
23. The value of c is 6 ⋅ 6 = 36; x2 + 12x + 36 = (x + 6)2.
24. The value of c is 10 ⋅ 10 = 100; x2 + 20x + 100 = (x + 10)2.
25. x2 + 6x + 3 = 0 x2 + 6x = −3
x2 + 6x + 9 = −3 + 9 (x + 3)2 = 6 x + 3 = ± √
— 6
x = −3 ± √—
6
The solutions are x = −3 − √—
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
The solutions are x = −2 − √—
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
The solutions are x = 4 − √—
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.
Copyright © Big Ideas Learning, LLC Algebra 2 111All rights reserved. Worked-Out Solutions
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
The solutions are x = 7 and x = −3.
112 Algebra 2 Copyright © Big Ideas Learning, LLC Worked-Out Solutions All rights reserved.
Chapter 3
42. Use factoring because the left-hand side factors.
x2 + 13x + 22 = 0 (x + 11)(x + 2) = 0 x + 11 = 0 or x + 2 = 0 x = −11 or x = −2
The solutions are x = −11 and 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.
45. Use factoring because the left-hand side factors.
x2 + 12x + 36 = 0 (x + 6)2 = 0 x + 6 = 0 x = −6
The solution is x = −6.
46. Use factoring because the left-hand side factors.
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
The solutions are x = −1 − √—
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
The solutions are x = −2 − √—
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
The solutions are x = −10 and 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
The solutions are x = − √—
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 − √—
57 , because lengths are
positive. So, x = −3 + √—
57 .
Copyright © Big Ideas Learning, LLC Algebra 2 113All rights reserved. Worked-Out Solutions
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 − √—
11 , because lengths are
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 ) .
114 Algebra 2 Copyright © Big Ideas Learning, LLC Worked-Out Solutions All rights reserved.
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.
64. Write the function in vertex form by completing the square.
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.
Copyright © Big Ideas Learning, LLC Algebra 2 115All rights reserved. Worked-Out Solutions
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.
116 Algebra 2 Copyright © Big Ideas Learning, LLC Worked-Out Solutions All rights reserved.
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.
Maintaining Mathematical Profi ciency
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.
Copyright © Big Ideas Learning, LLC Algebra 2 117All rights reserved. Worked-Out Solutions
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
The solutions are x = − √—
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 .
7. Set the real parts equal to each other and the imaginary parts
equal to each other.
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.
118 Algebra 2 Copyright © Big Ideas Learning, LLC Worked-Out Solutions All rights reserved.
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.
17. The resistor has a resistance of 7 ohms, so its impedance
is 7 ohms. The inductor has a reactance of 5 ohms, so its
impedance is 5i ohms. The capacitor has a reactance of
2 ohms, so its impedance is −2i ohms.
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.
3.4 Explorations (p. 121)
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.
Copyright © Big Ideas Learning, LLC Algebra 2 119All rights reserved. Worked-Out Solutions
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.
3.4 Monitoring Progress (pp. 122–126)
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 .
120 Algebra 2 Copyright © Big Ideas Learning, LLC Worked-Out Solutions All rights reserved.
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.
Solutions: x = −b ± √
— 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
The equation has two real solutions.
Solutions: x = −b ± √
— b2 − 4ac ——
2a
= −(−3) ± √
— 177 ——
2(7)
= 3 ± √
— 177 —
14
11. Equation: 4x2 + 6x + 9 = 0
Discriminant: b2 − 4ac = 62 − 4(4)(9) = −108
The equation has two imaginary solutions.
Solutions: x = −b ± √
— b2 − 4ac ——
2a
= −6 ± √
— −108 ——
2(4)
= −6 ± 6i √
— 3 —
8
= −3 ± 3i √
— 3 —
4
Copyright © Big Ideas Learning, LLC Algebra 2 121All rights reserved. Worked-Out Solutions
Chapter 3
12. Equation: −5x2 + 10x − 5 = 0
Discriminant: b2 − 4ac = 102 − 4(−5)(−5) = 0
The equation has one real solution.
Solution: x = −b ± √
— 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.
3.4 Exercises (pp. 127–130)
Vocabulary and Core Concept Check
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.
Monitoring Progress and Modeling with Mathematics
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
So, the solution is 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
122 Algebra 2 Copyright © Big Ideas Learning, LLC Worked-Out Solutions All rights reserved.
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
Copyright © Big Ideas Learning, LLC Algebra 2 123All rights reserved. Worked-Out Solutions
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
So, the solution is 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
Discriminant: b2 − 4ac = 122 − 4(1)(36) = 0
The equation has one real solution.
Solution: x = −b ± √
— b2 − 4ac ——
2a =
−12 ± √—
0 —
2(1) = −6
124 Algebra 2 Copyright © Big Ideas Learning, LLC Worked-Out Solutions All rights reserved.
Chapter 3
20. Equation: x2 − x + 6 = 0
Discriminant: b2 − 4ac = (−1)2 − 4(1)(6) = −23
The equation has two imaginary solutions.
Solutions: x = −b ± √
— 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
The equation has two real solutions.
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
The equation has two real solutions.
Solutions: x = −b ± √
— b2 − 4ac ——
2a
x = −2 ± √
— 52 —
2(1)
x = −1 ± √—
13
23. Equation: 4x2 − 5x + 10 = 0
Discriminant: b2 − 4ac = (−5)2 − 4(4)(10) = −135
The equation has two imaginary solutions.
Solutions: x = −b ± √
— b2 − 4ac ——
2a
x = −(−5) ± √
— −135 ——
2(4)
x = 5 ± 3i √
— 15 —
8
24. Equation: p2 + 18p + 81 = 0
Discriminant: b2 − 4ac = 182 − 4(1)(81) = 0
The equation has one real solution.
Solution: p = −b ± √
— b2 − 4ac ——
2a =
−18 ± √—
0 —
2(1) = −9
25. Equation: 3x2 + 24x + 48 = 0
Discriminant: b2 − 4ac = 242 − 4(3)(48) = 0
The equation has one real solution.
Solution: x = −b ± √
— b2 − 4ac ——
2a =
−24 ± √—
0 —
2(3) = −4
26. Equation: −2x2 − x − 6 = 0
Discriminant: b2 − 4ac = (−1)2 − 4(−2)(−6) = −47
The equation has two imaginary solutions.
Solutions: x = −b ± √
— 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
Copyright © Big Ideas Learning, LLC Algebra 2 125All rights reserved. Worked-Out Solutions
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.
36. 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 = 24
62 − 4ac = 24
36 − 4ac = 24
−4ac = −12
ac = 3
Because ac = 3, choose two integers whose product is 3,
such as a = −3 and c = −1. So, one possible equation is
−3x2 + 6x − 1 = 0.
37. 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 = 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
Because ac = 9, choose two integers whose product is 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
Because ac = 8, choose two integers whose product is 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.
126 Algebra 2 Copyright © Big Ideas Learning, LLC Worked-Out Solutions All rights reserved.
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.
48. Quadratic Formula:
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.
Copyright © Big Ideas Learning, LLC Algebra 2 127All rights reserved. Worked-Out Solutions
Chapter 3
49. Quadratic Formula:
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.
50. Quadratic Formula:
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.
51. Quadratic Formula:
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.
52. Quadratic Formula:
x2 + 8x − 13 = 0
x = −8 ± √——
82 − 4(1)(−13) ——
2(1)
x = −8 ± √—
116 —
2
x = −8 ± 2 √—
29 —
2
x = −4 ± √—
29
Completing the square:
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.
128 Algebra 2 Copyright © Big Ideas Learning, LLC Worked-Out Solutions All rights reserved.
Chapter 3
53. Quadratic Formula:
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
Completing the square:
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.
54. Quadratic Formula:
8x2 + 4x + 5 = 0
x = −4 ± √——
42 − 4(8)(5) ——
2(8)
x = −4 ± √—
−144 ——
16
x = −4 ± 12i —
16
x = −1 ± 3i —
4
Completing the square:
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.
55. Quadratic Formula:
−3 = 4x2 + 9x
4x2 + 9x + 3 = 0
x = −9 ± √——
92 − 4(4)(3) ——
2(4)
x = 9 ± √—
33 —
8
Completing the square:
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.
56. Quadratic Formula:
−31x + 56 = −x2
x2 + 31x + 56 = 0
x = −31 ± √——
312 − 4(1)(56) ——
2(1)
x = −31 ± √—
737 ——
2
Completing the square:
−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.
Copyright © Big Ideas Learning, LLC Algebra 2 129All rights reserved. Worked-Out Solutions
Chapter 3
57. Quadratic Formula:
x2 = 1 − x x2 + x − 1 = 0
x = −1 ± √——
12 − 4(1)(−1) ——
2(1)
x = −1 ± √—
5 —
2
Completing the square:
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.
58. Quadratic Formula:
9x2 + 36x + 72 = 0
x = −36 ± √——
362 − 4(9)(72) ——
2(9)
x = −36 ± √—
−1296 ——
18
x = −36 ± 36i —
18
x = −2 ± 2i
Completing the square:
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,
solve for t when h = 3.
h = −16t2 + v0t + h0
3 = −16t2 + 90t + 7 0 = −16t2 + 90t + 4 This equation is not factorable, and completing the square
would result in fractions. So, use the Quadratic Formula to
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.
130 Algebra 2 Copyright © Big Ideas Learning, LLC Worked-Out Solutions All rights reserved.
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,
solve for t when h = 0.
h = −16t2 + v0t + h0
0 = −16t2 − 55t + 10
This equation is not factorable, and completing the square
would result in fractions. So, use the Quadratic Formula to
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
This equation is not factorable, and completing the square
would result in fractions. So, use the Quadratic Formula to
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.
Copyright © Big Ideas Learning, LLC Algebra 2 131All rights reserved. Worked-Out Solutions
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
This equation is not factorable, and completing the square
would result in fractions. So, use the Quadratic Formula to
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
This equation is not factorable, and completing the square
would result in fractions. So, use the Quadratic Formula to
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.
132 Algebra 2 Copyright © Big Ideas Learning, LLC Worked-Out Solutions All rights reserved.
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.
Copyright © Big Ideas Learning, LLC Algebra 2 133All rights reserved. Worked-Out Solutions
Chapter 3
Maintaining Mathematical Profi ciency
77.
x
y
4
2
4 62
(4, 5)
The solution is (4, 5).
78.
x
y
4
6
2
42
(2, 3)
The solution is (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)
The solution is (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 )
134 Algebra 2 Copyright © Big Ideas Learning, LLC Worked-Out Solutions All rights reserved.
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).
Copyright © Big Ideas Learning, LLC Algebra 2 135All rights reserved. Worked-Out Solutions
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
Because x is an imaginary number, the system does not have
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)
The solutions are x = 2 and x = −2.
3.5 Exercises (pp.136–138)
Vocabulary and Core Concept Check
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.
Monitoring Progress and Modeling with Mathematics
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)
The solution is (3, 5).
Check:
x = 3:
y = (x − 3)2 + 5 = (3 − 3)2 + 5 = 5 y = 5
136 Algebra 2 Copyright © Big Ideas Learning, LLC Worked-Out Solutions All rights reserved.
Chapter 3
5.
x
y
4−4
4
The graphs do not intersect, so there is no solution.
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
The graphs do not intersect, so there is no solution.
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)
The solutions are (−2, 0) and (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).
Copyright © Big Ideas Learning, LLC Algebra 2 137All rights reserved. Worked-Out Solutions
Chapter 3
15. Substitute x + 5 for y in Equation 2 and solve for x.
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
Because x is an imaginary number, the system does not have
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).
138 Algebra 2 Copyright © Big Ideas Learning, LLC Worked-Out Solutions All rights reserved.
Chapter 3
22. Begin by solving for y in Equation 1.
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
Because x is an imaginary number, the system does not have
a real solution.
24. Begin by solving for y in Equation 2.
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
To solve for y, substitute x = 1 and x = 8 into the equation
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
Because x is an imaginary number, the system does not have
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).
Copyright © Big Ideas Learning, LLC Algebra 2 139All rights reserved. Worked-Out Solutions
Chapter 3
28. Add the equations to eliminate the y-term and obtain a
quadratic equation in x.
−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.
29. Add the equations to eliminate the y-term and obtain a
quadratic equation in x.
−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
quadratic equation in x.
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).
31. Add the equations to eliminate the y-term and obtain a
quadratic equation in x.
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 ).
32. Add the equations to eliminate the y-term and obtain a
quadratic equation in x.
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
The solutions are (−4 − √—
2 , 9 + 4 √—
2 ) and
(−4 + √—
2 , 9 − 4 √—
2 ).
140 Algebra 2 Copyright © Big Ideas Learning, LLC Worked-Out Solutions All rights reserved.
Chapter 3
33. Subtract the equations to eliminate the y-term and obtain a
quadratic equation in x.
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).
34. Subtract the equations to eliminate the y-term and obtain a
quadratic equation in x.
−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
The solution is (4, −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
quadratic equation in x.
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
The solution is (0, −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.
39. Use the substitution method because one equation is solved
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
The solutions are (−3 − √—
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
quadratic equation in x.
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).
Copyright © Big Ideas Learning, LLC Algebra 2 141All rights reserved. Worked-Out Solutions
Chapter 3
41. Use the substitution method because one equation is solved
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
To solve for y, substitute x = 3 and x = 5 into the equation
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).
42. Use the substitution method because one equation is solved
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.
44. Write a system of equations using each side of the original
equation.
Equation System
2x2 − 12x − 16 = y = 2x2 − 12x − 16
−6x2 + 60x − 144 y = −6x2 + 60x − 144
Graph the equations in the same plane.
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.
45. Write a system of equations using each side of the original
equation.
Equation System
(x + 2)(x − 2) = −x2 + 6x − 7 y = (x + 2)(x − 2)
y = −x2 + 6x − 7
Graph the equations in the same plane.
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.
46. Write a system of equations using each side of the original
equation.
Equation System
−2x2 − 16x − 25 = 6x2 + 48x + 95 y = −2x2 − 16x − 25
y = 6x2 + 48x + 95
Graph the equations in the same plane.
x
y
3
7
1
−1 1−5−7
(−5, 5) (−3, 5)
The solutions are x = −5 and x = −3.
142 Algebra 2 Copyright © Big Ideas Learning, LLC Worked-Out Solutions All rights reserved.
Chapter 3
47. Write a system of equations using each side of the original
equation.
Equation System
(x − 2)2 − 3 = y = (x − 2)2 − 3
(x + 3)(−x + 9) − 38 y = (x + 3)(−x + 9) − 38
Graph the equations in the same plane.
x
y
−4
−2
42
(3, −2)
(2, −3)
The solutions are x = 2 and x = 3.
48. Write a system of equations using each side of the original
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
Graph the equations in the same plane.
x
y
−4
−6
−2
−2−4
The graphs do not intersect, so there is no solution.
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.
Copyright © Big Ideas Learning, LLC Algebra 2 143All rights reserved. Worked-Out Solutions
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
The solutions are (−1, 1) and (1, 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
144 Algebra 2 Copyright © Big Ideas Learning, LLC Worked-Out Solutions All rights reserved.
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.
Copyright © Big Ideas Learning, LLC Algebra 2 145All rights reserved. Worked-Out Solutions
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.
Maintaining Mathematical Profi ciency
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.
3.6 Explorations (p. 139)
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.
146 Algebra 2 Copyright © Big Ideas Learning, LLC Worked-Out Solutions All rights reserved.
Chapter 3
3.6 Monitoring Progress (pp. 141–143)
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
So, (2, 0) is a solution of the inequality.
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.
Step 2 Test a point outside the parabola, such as (0, 5).
y > −x2 + 2x + 4
5 >? −02 + 2(0) + 4
5 > 4
So, (0, 5) is a solution of the inequality.
Step 3 Shade the region outside the parabola.
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
Copyright © Big Ideas Learning, LLC Algebra 2 147All rights reserved. Worked-Out Solutions
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 .
7. First, write and solve the equation obtained by replacing
> 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.
148 Algebra 2 Copyright © Big Ideas Learning, LLC Worked-Out Solutions All rights reserved.
Chapter 3
3.6 Exercises (pp. 144–146)
Vocabulary and Core Concept Check
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.
Monitoring Progress and Modeling with Mathematics
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)
satisfi es the inequality.
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.
Step 3 Shade the region inside the parabola.
y
42−2−4
−4
−2
2
4
x
8. Step 1 Graph y = 4x2. Because the inequality symbol is ≥,
make the parabola solid.
Step 2 Test a point inside the parabola, such as (0, 1).
y ≥ 4x2
1 ≥? 4(0)2
1 ≥ 0
So, (0, 1) is a solution of the inequality.
Step 3 Shade the region inside the parabola.
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.
Step 2 Test a point inside the parabola, such as (0, 1).
y > x2 − 9
1 ≥? (0)2 − 9
1 ≥ −9
So, (0, 1) is a solution of the inequality.
Step 3 Shade the region inside the parabola.
y
4 62−2−4−6
−4
−6
−10
−2
2
x
Copyright © Big Ideas Learning, LLC Algebra 2 149All rights reserved. Worked-Out Solutions
Chapter 3
10. Step 1 Graph y = x2 + 5. Because the inequality symbol
is <, make the parabola dashed.
Step 2 Test a point outside the parabola, such as (0, 1).
y < x2 + 5
1 <? (0)2 + 5
1 < 5
So, (0, 1) is a solution of the inequality.
Step 3 Shade the region outside the parabola.
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.
Step 3 Shade the region outside the parabola.
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.
Step 2 Test a point outside the parabola, such as (0, 0).
y ≥ −2x2 + 9x − 4
0 ≥? −2(0)2 + 9(0) − 4
0 ≥ −4
So, (0, 0) is a solution of the inequality.
Step 3 Shade the region outside the parabola.
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.
Step 3 Shade the region inside the parabola.
y
−4−6
−2
2
4
6
8
x
150 Algebra 2 Copyright © Big Ideas Learning, LLC Worked-Out Solutions All rights reserved.
Chapter 3
14. Step 1 Graph y = ( x − 1 —
2 ) 2 +
5 —
2 . Because the inequality
symbol is ≤, make the parabola solid.
Step 2 Test a point outside the parabola, such as (0, 0).
y ≤ ( x − 1 —
2 )
2
+ 5 —
2
0 ≤? ( 0 −
1 —
2 )
2
+ 5 —
2
0 ≤ 11
— 4
So, (0, 0) is a solution of the inequality.
Step 3 Shade the region outside the parabola.
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.
Step 3 Identify the region where the two graphs overlap.
This region is the graph of the system.
y
2−2
−2
2
x
y ≥ 2x2
y < −x2 + 1
Copyright © Big Ideas Learning, LLC Algebra 2 151All rights reserved. Worked-Out Solutions
Chapter 3
22. Step 1 Graph y > −5x2.
Step 2 Graph y > 3x2 − 2.
Step 3 Identify the region where the two graphs overlap.
This region is the graph of the system.
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.
Step 3 Identify the region where the two graphs overlap.
This region is the graph of the system.
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.
Step 3 Identify the region where the two graphs overlap.
This region is the graph of the system.
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.
Step 3 Identify the region where the two graphs overlap.
This region is the graph of the system.
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.
Step 3 Identify the region where the two graphs overlap.
This region is the graph of the system.
y
2 4 6−4−8 −2
6
8
x
−6
−8 y ≥ x2 − 3x − 6y ≥ x2 + 7x + 6
152 Algebra 2 Copyright © Big Ideas Learning, LLC Worked-Out Solutions All rights reserved.
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
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.
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 .
28. First, write and solve the equation obtained by replacing
< 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
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 = −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
x-values partition the number line into three intervals. Test an
x-value in each interval to determine whether it satisfi es the
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
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 = 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 .
Copyright © Big Ideas Learning, LLC Algebra 2 153All rights reserved. Worked-Out Solutions
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
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 = −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
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 = −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
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 = −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 .
154 Algebra 2 Copyright © Big Ideas Learning, LLC Worked-Out Solutions All rights reserved.
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
36. The solution consists of the x-values for which the graph of
y = x2 − 4x + 2 lies above the x-axis. Find the x-intercepts
of the graph by letting y = 0 and use the Quadratic Formula
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
38. The solution consists of the x-values for which the graph of
y = x2 + 6x + 3 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 + 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
Copyright © Big Ideas Learning, LLC Algebra 2 155All rights reserved. Worked-Out Solutions
Chapter 3
39. The solution consists of the x-values for which the
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
40. The solution consists of the x-values for which the
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
41. The solution consists of the x-values for which the
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
156 Algebra 2 Copyright © Big Ideas Learning, LLC Worked-Out Solutions All rights reserved.
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.
Copyright © Big Ideas Learning, LLC Algebra 2 157All rights reserved. Worked-Out 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.
158 Algebra 2 Copyright © Big Ideas Learning, LLC Worked-Out Solutions All rights reserved.
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.
Maintaining Mathematical Profi ciency
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
Next, fi nd the y-coordinate of the vertex.
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.
Copyright © Big Ideas Learning, LLC Algebra 2 159All rights reserved. Worked-Out Solutions
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
Next, fi nd the y-coordinate of the vertex.
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)
1. The equation is in standard form. Graph the related function
y = x2 − 2x − 8.
x
y4
−8
−4
The x-intercepts are −2 and 4. The solutions, or roots, are
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.
6. Set the real parts equal to each other and the imaginary parts
equal to each other.
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 .
160 Algebra 2 Copyright © Big Ideas Learning, LLC Worked-Out Solutions All rights reserved.
Chapter 3
11. 2x2 + 32 = 0 2x2 = −32
x2 = −16
x = ± √—
−16
x = ±4i
So, the zeros of f are 4i and −4i.
12. Write the function in vertex form by completing the square.
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
The solutions are x = −8 − √—
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.
Solution: x = −b ± √
— b2 − 4ac ——
2a =
−(−6) ± √—
0 ——
2(−1) = −3
Copyright © Big Ideas Learning, LLC Algebra 2 161All rights reserved. Worked-Out Solutions
Chapter 3
21. Equation: x2 − 2x − 9 = 0 Discriminant: b2 − 4ac = (−2)2 − 4(1)(−9) = 40
The equation has two real solutions.
Solutions: x = −b ± √
— 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
The equation has two real solutions.
Solutions: x = −b ± √
— 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.
162 Algebra 2 Copyright © Big Ideas Learning, LLC Worked-Out Solutions All rights reserved.
Chapter 3
27. Step 1 Graph y = x2 + 8x + 16. Because the inequality
symbol is >, make the parabola dashed.
Step 2 Test a point inside the parabola, such as (−4, 1).
y > x2 + 8x + 16
1 >? (−4)2 + 8(−4) + 16
1 > 0 So, (−4, 1) is a solution of the inequality.
Step 3 Shade the region inside the parabola.
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.
Step 2 Test a point inside the parabola, such as (−3, 1).
y ≥ x2 + 6x + 8 1 ≥
? (−3)2 + 6(−3) + 8
1 ≥ −1
So, (−3, 1) is a solution of the inequality.
Step 3 Shade the region inside the parabola.
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.
Step 4 Shade the region inside the parabola.
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.
Step 3 Identify the region where the two graphs overlap.
This region is the graph of the system.
x
y
6
4
2
8
42−2 6
y < x2 − 4x + 8
y ≥ −x2 + 4x + 2
Copyright © Big Ideas Learning, LLC Algebra 2 163All rights reserved. Worked-Out Solutions
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.
Step 4 Identify the region where the two graphs overlap.
This region is the graph of the system.
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.
Step 4 Identify the region where the two graphs overlap.
This region is the graph of the system.
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
partition the number line into three intervals. Test an x-value in
each interval to determine whether it satisfi es the inequality.
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
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 = −4.
−3 −2 −1 0−4−5−6−7−8−9−10
Test x = 0.Test x = −8.
−(−8)2 − 10(−8) < 21 ✓
−(−4)2 − 10(−4) ≮ 21
−(0)2 − 10(0) < 21 ✓
So, the solution is x < −7 or x > −3.
164 Algebra 2 Copyright © Big Ideas Learning, LLC Worked-Out Solutions All rights reserved.
Chapter 3
35. First, write and solve the equation obtained by replacing
≤ 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
x-values partition the number line into three intervals. Test an
x-value in each interval to determine whether it satisfi es the
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
The solutions are x = 3 − √—
2 and x = 3 + √—
2 .
3. Use the square root method because the equation can be
written in the form u2 = d.
x2 + 49 = 85
x2 = 36
x = ±6
The solutions are x = −6 and x = 6.
4. Use the square root method because the equation can be
written in the form u2 = d.
(x + 4)( x − 1) = −x2 + 3x + 4 x2 + 3x − 4 = −x2 + 3x + 4 2x2 = 8 x2 = 4 x = ±2
The solutions are x = −2 and 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.
8. Begin by solving for y in Equation 2.
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).
Copyright © Big Ideas Learning, LLC Algebra 2 165All rights reserved. Worked-Out Solutions
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
10. Substitute x + 4 for y in Equation 1 and solve for x.
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
Next, fi nd the y-coordinate of the vertex.
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.
166 Algebra 2 Copyright © Big Ideas Learning, LLC Worked-Out Solutions All rights reserved.
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.