1. Exploring Quadratic Graphs
ALGEBRA 1 LESSON 10-1
(For help, go to Lessons 1-2 and 5-3.)
Evaluate each expression for h = 3, k = 2, and j = –4.
1. hkj 2. kh hk kj h
Graph each equation.
5. y = 2x – y = | x | 7. y = x
10-1

2. Exploring Quadratic Graphs
ALGEBRA 1 LESSON 10-1
1. hkj for h = 3, k = 2, j = –4: (3)(2)(–4) = 6(–4) = –24
2. kh2 for h = 3, k = 2: 2(32) = 2(9) = 18
3. hk2 for h = 3, k = 2: 3(22) = 3(4) = 12
4. kj 2 + h for h = 3, k = 2, j = –4: 2(–4)2 + 3 = 2(16) + 3 = = 35
5. y = 2x – y = | x |
7. y = x
Solutions
10-1

3. Exploring Quadratic Graphs
ALGEBRA 1 LESSON 10-1
Identify the vertex of each graph. Tell whether it is a minimum or a maximum. a. b.
The vertex is (1, 2).
The vertex is (2, –4).
It is a maximum.
It is a minimum.
10-1

4. Exploring Quadratic Graphs
ALGEBRA 1 LESSON 10-1
Make a table of values and graph the quadratic
function y = x2. 1 3
x y = x2 (x, y) 1 3
0 (0)2 = (0, 0)
2 (2)2 = 1 (2, 1 )
3 (3)2 = (3, 3)
10-1

5. Exploring Quadratic Graphs
ALGEBRA 1 LESSON 10-1
Use the graphs below. Order the quadratic functions
(x) = –x2, (x) = –3x2, and (x) = x2 from widest to narrowest graph. 1 2
(x) = x2 1 2
(x) = –x2
(x) = –3x2
Of the three graphs, (x) = x2 is the widest and (x) = –3x2 is the narrowest. 1 2
So, the order from widest to narrowest is (x) = x2, (x) = –x2,
(x) = –3x2. 1 2
10-1

6. Exploring Quadratic Graphs
ALGEBRA 1 LESSON 10-1
Graph the quadratic functions y = 3x2 and y = 3x2 – 2. Compare the graphs.
x y = 3x2 y = 3x2 – 2 2 –
The graph of y = 3x2 – 2 has the same shape as the graph of y = 3x2, but it is shifted down 2 units.
10-1

7. Exploring Quadratic Graphs
ALGEBRA 1 LESSON 10-1
A monkey drops an orange from a branch 26 ft above the ground. The force of gravity causes the orange to fall toward Earth. The function h = –16t gives the height of the orange, h, in feet after t seconds. Graph this quadratic function.
Use positive values for t.
Graph t on the x-axis and h on the y-axis.
Height h is dependent on time t.
t h = –16t2 + 26
0 26
1 10
2 –38
10-1

8. Exploring Quadratic Graphs
ALGEBRA 1 LESSON 10-1
pages 513–516  Exercises
1. (2, 5); max.
2. (–3, –2); min.
3. (2, 1); min. 4. 5. 6. 7. 8. 9.
10. y = x2, y = 3x2, y = 4x2
11. ƒ(x) = x2, ƒ(x) = x2, ƒ(x) = 5x2
12. y = – x2, y = – x2, y = 5x2
13. ƒ(x) = – x2, ƒ(x) = –2x2,
ƒ(x) = –4x2
14. 1 2 1 3 1 4 1 2 2 3
10-1

9. Exploring Quadratic Graphs
ALGEBRA 1 LESSON 10-1
21. E
22. A
23. F
24. B
25. C
26. D
27. The graph of y = 2x2
is narrower.
28. The graph of y = –x2
opens downward.
29. The graph of y = 1.5x2
30. The graph of y = x2
is wider.
15.
16.
17.
18.
19.
20. 1 2
10-1

10. Exploring Quadratic Graphs
ALGEBRA 1 LESSON 10-1
31.
32.
33.
34.
35.
36.
37.
38. a.
b. 184 ft
c. 56 ft
39. a. 0 < r < 6
b. 0 < A < c.
10-1

11. Exploring Quadratic Graphs
ALGEBRA 1 LESSON 10-1
40. K, L
41. M
42. K
43. M
44. Answers may vary. Sample:
a. y = 5x2 b. y = –5x2 c. y = 3x2
45. a.
b. 16 ft
c. No; the apple falls 48 ft
from t = 1 to t = 2,
because it is accelerating.
46. a. c and a and c have opp. signs.
b. c and a and c have the same signs.
47. a.
b. 0 < x < 12; the side length of the square
garden must be less than the width
of the patio.
c. 96 < A < 240; as the side length of the
garden increases from 0 to 12, the area
of the patio decreases from 240 to 96.
d. about 6 ft = / = /
10-1

12. Exploring Quadratic Graphs
ALGEBRA 1 LESSON 10-1
50. B
51. G
52. D
53. [4] a.
b. 3.5 s
[3] estimate incorrect or missing
[2] error in table or graph
[1] table OR reasonable graph only
48. a. a > 0
b. |a| > 1
49. a. b.
t h
0 200
1 184
2 136
3   56
4 –56
10-1

13. Exploring Quadratic Graphs
ALGEBRA 1 LESSON 10-1
54. (x2 + 2)(x – 4)
55. (3a2 – 2)(5a – 6)
56. (7b2 + 1)(b + 2)
57. (y + 2)(y – 2)(y + 3)
58. 2n(n + 3)(n – 4)
59. 3m(2m + 3)(5m + 1)
60. 15x2 – 20x
61. 9n2 – 63n
62. –12t t 2
63. 12m6 – 4m5 + 20m2
64. –15y6 – 10y4 + 20y
65. –12c5 + 21c4 – 24c3
balloons
10-1

14. Exploring Quadratic Graphs
ALGEBRA 1 LESSON 10-1
1. a.  Graph y = – x2 – 1.
b.  Identify the vertex. Tell whether it is a maximum or a minimum.
c.  Compare this graph to the graph of y = x2.
2. a.  Graph y = 4x2 + 3.
b. Identify the vertex. Tell whether it is a maximum or a minimum. 1 2
(0, -1); maximum
This graph is wider, opens downward, and is shifted one unit down.
(0, 3); minimum
This graph is narrower and shifted 3 units up.
3. Order the quadratic functions y = –4x2, y = x2, and y = 2x2 from widest to narrowest graph. 1 4
y = x2, y = 2x2, y = –4x2 1 4
10-1

15. Evaluate the expression for the following values of a and b.
Quadratic Functions
ALGEBRA 1 LESSON 10-2
(For help, go to Lessons 1-6 and 10-1.)
Evaluate the expression for the following values of a and b.
1. a = –6, b = 4 2. a = 15, b = 20
3. a = –8, b = –56 4. a = –9, b = 108
Graph each function.
5. y = x2 6. y = –x y = x2 – 1 –b 2a 1 2
10-2

16. Quadratic Functions Solutions – for a = –6, b = 4: = 1. –
ALGEBRA 1 LESSON 10-2
Solutions – b 2a
for a = –6, b = 4: –4
2(–6) =
–12 1 3 1. – b 2a
for a = 15, b = 20:
–20
2(15)
–20 30 2 3 2. = = – – b 2a
for a = –8, b = –56:
–(–56)
2(–8) 56
–16 7 2 1 2 3. = = –
= –3 – b 2a
for a = –9, b = 108:
–108
2(–9) =
–18 6 4.
7. y = x2 – 1 1 2 5.
y = x2 6.
y = –x2 + 2
10-2

17. Graph the function y = 2x2 + 4x – 3.
Quadratic Functions
ALGEBRA 1 LESSON 10-2
Graph the function y = 2x2 + 4x – 3.
Step 1: Find the equation of the axis of symmetry and the
coordinates of the vertex.
Find the equation of the axis of symmetry.
x = b 2a – = –4
2(2)
= – 1
The x-coordinate of the vertex is –1.
y = 2x2 + 4x – 3
y = 2(–1)2 + 4(–1) – 3
= –5
To find the y-coordinate of the vertex, substitute –1 for x.
The vertex is (–1, –5).
10-2

18. Step 2: Find two other points. Use the y-intercept.
Quadratic Functions
ALGEBRA 1 LESSON 10-2
(continued)
Step 2: Find two other points.
Use the y-intercept.
For x = 0, y = –3, so one point is (0, –3).
Choose a value for x on the same side of the vertex.
Let x = 1
y = 2(1)2 + 4(1) – 3  
= 3
For x = 1, y = 3, so another point is (1, 3).
Find the y-coordinate for x = 1.
10-2

19. Quadratic Functions (continued)
ALGEBRA 1 LESSON 10-2
(continued)
Step 3: Reflect (0, –3) and (1, 3) across the axis of symmetry to get two more points.
Then draw the parabola.
10-2

20. Quadratic Functions
ALGEBRA 1 LESSON 10-2
Aerial fireworks carry “stars,” which are made of a sparkler-like material, upward, ignite them, and project them into the air in fireworks displays. Suppose a particular star is projected from an aerial firework at a starting height of 610 ft with an initial upward velocity of 88 ft/s. How long will it take for the star to reach its maximum height? How far above the ground will it be?
The equation h = –16t2 + 88t gives the height of the star h in feet at time t in seconds.
Since the coefficient of t 2 is negative, the curve opens downward, and the vertex is the maximum point.
10-2

21. Step 1: Find the x-coordinate of the vertex.
Quadratic Functions
ALGEBRA 1 LESSON 10-2
(continued)
Step 1:  Find the x-coordinate of the vertex.
After 2.75 seconds, the star will be at its greatest height. b 2a – =
–88
2(–16)
2.75
Step 2:  Find the h-coordinate of the vertex.
h = –16(2.75)2 + 88(2.75) Substitute 2.75 for t.
h = Simplify using a calculator.
The maximum height of the star will be about 731 ft.
10-2

22. Graph the quadratic inequality y > –x2 + 6x – 5.
Quadratic Functions
ALGEBRA 1 LESSON 10-2
Graph the quadratic inequality y > –x2 + 6x – 5.
Graph the boundary curve, y = –x2 + 6x – 5.
Use a dashed line because the solution of the inequality y > –x2 + 6x – 5 does not include the boundary.
Shade above the curve.
10-2

23. Quadratic Functions pages 520–523 Exercises 1. x = 0, (0, 4)
ALGEBRA 1 LESSON 10-2
pages 520–523  Exercises
1. x = 0, (0, 4)
2. x = –1, (–1, –7)
3. x = 4, (4, –25)
4. x = 1.5, (1.5, –1.75)
5. B
6. E
7. C
8. F
9. A
10. D
11.
12.
13.
14.
15. a. 20 ft
b. 400 ft2
16. a s
b. 31 ft
17.
10-2

24. Quadratic Functions
ALGEBRA 1 LESSON 10-2
18.
19.
20.
21.
22.
23.
24.
25.
26.
10-2

25. Quadratic Functions 27. 28. 29. 30. 31. 32–34. Answers may vary.
ALGEBRA 1 LESSON 10-2
27.
28.
29.
30.
31.
32–34. Answers may vary.
Samples are given.
32. y = 2x2 – 8x + 1
33. y = –3x2
34. y = 2x2 + 4
35. a. 1.2 m
b. 7.2 m
36. a. y < –0.1x2 + 12 b.
c. Yes; when x = 6, y = 8.4,
so the camper will fit.
37. a. $12.50
b. $10,000
units2
units2
10-2

26. 40. Answers may vary. Sample: If the coefficient of the squared
Quadratic Functions
ALGEBRA 1 LESSON 10-2
40. Answers may vary. Sample: If
the coefficient of the squared
term is pos., the vertex point is
a min.; if it is neg., the vertex
point is a max.
41. Answers may vary. Sample: a
affects whether the parabola
opens up or down, b affects
the axis of symmetry, and c
affects the y-intercept.
42. a. w = 13 –
b. A = –
c. (6.5, 42.25)
d. 6.5 ft by 6.5 ft
43. a. 0.4 s
b. No; after 0.6 s, the ball will
have a height of about 2.23 m
but the net has a height of 2.43 m.
44.
45. a. 0.4 s
b. No; it takes about 0.8 s to return to
h = 0.5 m, so it will take more time
to reach the ground.
10-2

27. [1] appropriate methods, but with a minor computational error 51. C
Quadratic Functions
ALGEBRA 1 LESSON 10-2
46. a. (0, 2)
b. x = –2.5
c. 5
d. y = x2 + 5x + 2
e. Answers may vary.
Sample: Test (–4, –2).
–2 (–4)2 + 5(–4) + 2
– –
–2 = –2
f. No; you would not be
able to determine the
b value using the
vertex formula.
47. A
48. I
49. B
50. [2] axis of symmetry:
x = =
maximum height:
y –0.009(16.7) (16.7) + 4.5
y ft
[1] appropriate methods, but with a minor computational error
51. C
52. A
53. F
54. D
55. B
56. E –b 2a
–0.3
2(–0.009)
10-2

28. Quadratic Functions 57. c2 – 5c – 36 58. 2x2 + 7x – 30
ALGEBRA 1 LESSON 10-2
57. c2 – 5c – 36
58. 2x2 + 7x – 30
59. 20t t + 3
60. 21n4 – 62n2 + 16
61. 2a3 + 9a2 – a + 20
62. 6r 3 + 9r 2 – 20r + 7
10-2

29. Graph each relation. Label the axis of symmetry and the vertex.
Quadratic Functions
ALGEBRA 1 LESSON 10-2
Graph each relation. Label the axis of symmetry and the vertex.
1. y = x2 – 8x + 15
2. ƒ(x) = –x2 + 4x – 2 1 4
<
3. y – x2 – 2x – 6
x = –4
10-2

30. Finding and Estimating Square Roots
ALGEBRA 1 LESSON 10-3
(For help, go to Lessons 1-2 and 8-5.)
Simplify each expression.
(–12)2 3. –(12) 1 2 3 – 4 5
10-3

31. Finding and Estimating Square Roots
ALGEBRA 1 LESSON 10-3
Solutions
= 11 • 11 = (–12)2 = (–12)(–12) = 144
3. (–12)2 = –(12)(12) = – = (1.5)(1.5) = 2.25
= (0.6)(0.6) = = • =
= = = • = 1 2 1 2 1 2 1 4 2 3 – 2 3 – 2 3 – 4 9 4 5 4 5 4 5 16 25
10-3

32. Finding and Estimating Square Roots
ALGEBRA 1 LESSON 10-3
Simplify each expression. a
positive square root
= 5
b. ± 9 25 3 5
The square roots are and – .
= ±
c. – 64
negative square root
= –8
d. –49
For real numbers, the square root of a negative
number is undefined.
is undefined
e. ± 0
There is only one square root of 0.
= 0
10-3

33. Finding and Estimating Square Roots
ALGEBRA 1 LESSON 10-3
Tell whether each expression is rational or irrational.
a. ± = ± 12
rational
b. – = – 1 5
irrational
c. – = –2.5
rational
d = 0.3 1 9
rational
e =
irrational
10-3

34. Finding and Estimating Square Roots
ALGEBRA 1 LESSON 10-3
Between what two consecutive integers is ?
28.34 is between the two consecutive perfect squares 25 and 36.
25 < <
The square roots of 25 and 36 are 5 and 6, respectively.
28.34
5 < < 6
28.34 is between 5 and 6.
10-3

35. Finding and Estimating Square Roots
ALGEBRA 1 LESSON 10-3
Find to the nearest hundredth.
28.34
Use a calculator.
5.32
Round to the nearest hundredth.
10-3

36. Finding and Estimating Square Roots
ALGEBRA 1 LESSON 10-3
Suppose a rectangular field has a length three times its width x. The formula d = x2 + (3x)2 gives the distance of the diagonal of a rectangle. Find the distance of the diagonal across the field if x = 8 ft.
d = x2 + (3x)2
d = (3 • 8)2
Substitute 8 for x.
Simplify.
d =
d =
Use a calculator. Round to the nearest tenth. d
The diagonal is about 25.3 ft long.
10-3

37. Finding and Estimating Square Roots
ALGEBRA 1 LESSON 10-3
pages 526–528  Exercises
1. 13
2. 20 3.
4. 30
5. 0.5 6.
7. –1.1
8. 1.4
9. 0.6
10. –12
11.
12. ± 0.1
13. irrational
14. rational
15. irrational
16. rational
17. 5 and 6
18. 5 and 6
19. –12 and –11
and 14
22. –14.25
24. –12.25
26. ± 20
27. 0
28. ± 25
29. ±
30. ± 1.3
31. ±
32. ± 27
33. ± 1.5
34. ± 16
35. ± 0.1 1 3 3 7 6 7 1 9 5 4
10-3

38. Finding and Estimating Square Roots
ALGEBRA 1 LESSON 10-3
49. Answers may vary. Sample:
The first expression means
the neg. square root of 1 and
the second expression means
the pos. square root of 1.
50. Answers may vary.
Sample: 3 and 4
51. 4
52. a. 5 s
b. 10 s
c. No; the object takes
twice as long to fall.
53. False; zero has one square root.
54. false; = 1
55. true
56. true
57. False; answers may vary.
Sample:
58. False; answers may vary.
Sample: and are
irrational but is rational.
59. a. 4 units2
b. unit2
c. 2 units2
d units
36. ±
37. ± 202
38. 1
39. a km
b km
40. 21
41. –
44. –5.48
45. –33
46. –0.8 8 11 = / 2 5 1 2
10-3

39. Finding and Estimating Square Roots
ALGEBRA 1 LESSON 10-3
61. 6
62.
63. 9
64. 11
66.
67.
68.
69.
70.
71.
72. d 2 – 81
73. 9t 2 – 25 2 9
10-3

40. Finding and Estimating Square Roots
ALGEBRA 1 LESSON 10-3
75. x2 + 26x + 169
76. 16y2 – 56y + 49
77. 10,201
,604
79. 36k2 – 49
b2 – 168b + 49
81. –2
82.
83. –
84. –972
85.
86. 3 16 3 4 1 12 81 4
10-3

41. Finding and Estimating Square Roots
ALGEBRA 1 LESSON 10-3
1. Simplify each expression.
a         b.  ±
2. Tell whether each expression is rational, irrational, or undefined.
a. ±       b.  –25       c. –
3. Between what two consecutive integers is – 54?
4. The formula s = d estimates the speed s in miles per hour that a car was traveling, when it applied its brakes and left a skid mark d feet long on a wet road. Estimate the speed of a car that left a 120 foot long skid mark. 4 25 2 5 14 3 5
irrational
undefined
rational
–8 and –7
about mph
10-3

42. Solving Quadratic Equations
ALGEBRA 1 LESSON 10-4
(For help, go to Lesson10-3.)
Simplify each expression.
– ± ±
± 9. ± 1 4 9 49
100
10-4

43. Solving Quadratic Equations
ALGEBRA 1 LESSON 10-4
Solutions
= 6 2. – = –9
3. ± = ± = 1.2
= ± = ±1.1
= 8. ± = ±
9. ± = 1 4 1 2 1 9 1 3 49
100 7 10
10-4

44. Solving Quadratic Equations
ALGEBRA 1 LESSON 10-4
Solve each equation by graphing the related function.
a. 2x2 = 0
b. 2x2 + 2 = 0
c. 2x2 – 2 = 0
Graph y = 2x2
Graph y = 2x2 + 2
Graph y = 2x2 – 2
There is one solution, x = 0.
There is no solution.
There are two solutions, x = ±1.
10-4

45. Solving Quadratic Equations
ALGEBRA 1 LESSON 10-4
Solve 3x2 – 75 = 0.
3x2 – = Add 75 to each side.
3x2 = 75
x2 = 25 Divide each side by 3.
x = ± Find the square roots.
x = ± 5 Simplify.
10-4

46. Solving Quadratic Equations
ALGEBRA 1 LESSON 10-4
A museum is planning an exhibit that will contain a large globe. The surface area of the globe will be 315 ft2. Find the radius of the sphere producing this surface area. Use the equation S = 4 r2, where S is the surface area and r is the radius.
S = 4 r 2
315 = 4 r 2
Substitute 315 for S.
Put in calculator ready form.
315
(4)
= r 2
= r 2
315
(4)
Find the principle square root. r
Use a calculator.
The radius of the sphere is about 5 ft.
10-4

47. Solving Quadratic Equations
ALGEBRA 1 LESSON 10-4
pages 531–534  Exercises 1. ±3 2.
no solution 3. 4. ±2 5.
no solution 6. ±3 7.
no solution 8. 9. ±2
10-4

48. Solving Quadratic Equations
ALGEBRA 1 LESSON 10-4
10. ± 7
11. ± 21
12. ± 15
13. 0
14. no solution
15. ±
16. ±
17. ± 2
18. ±
19. x2 = 256; 16 m
20. x2 = 90; 9.5 ft
21. r 2 = 80; 5.0 cm
30. ±
31. ± 2.8
32. ± 0.4
33. ± 3.5 s
36. a. n > 0
b. n = 0
c. n < 0
37. Answers may vary.
Sample: Michael
subtracted 25 from
the left side of the
equation but added
25 to the right side. 1 6
22. a. 6.0 in.
b. The length of a radius
cannot be negative.
23. none
24. two
25. one
in. by 10.4 in.
27. a ft
b ft
c. No; the radius increases
by about 1.4 times.
28. no solution
29. ± 5 2 1 4 3 7
10-4

49. Solving Quadratic Equations
ALGEBRA 1 LESSON 10-4 ft cm
43. a. 0.2 m
b. 2.5 s
c. 3.0 s
d. Shorten; as decreases, t decreases.
44. a. –7
b. (–7, 0)
c. Answers may vary.
Sample: h = 5, –5, (–5, 0)
d. (4, 0); the vertex is at (–h, 0). cm
46. B
47. I
48. B
38. a. 2, –2; 2, –2
b. The first equation multiplied
by 2 on both sides equals
the second equation.
39. a. square: 4r 2, circle: r 2
b. 4r 2 – r 2 = 80
c. 9.7 in., 19.3 in.
40. Answers may vary. Sample:
a. 5x = 0, no solution
b. 2x2 + 0 = 0, x = 0
c. –20x = 0, x = ± 2
10-4

50. Solving Quadratic Equations
ALGEBRA 1 LESSON 10-4
49. [2]
x-intercepts
1.5, –1.5
[1] minor error in table
OR incorrect graph
50. [4] a. 96 = 6s2
s2 = 16
s = 4, so side is 4 ft.
b. 6(8)2 = 6 • 64 = 384,
so surface area is 384 ft2.
The surface area is quadrupled.
[3] appropriate methods, but with one computational error
[2] part (a) done correctly
[1] no work shown
51. 3
52. –13
53. 40
54. 15
x y
–2 5
–1 –4
0 –7
1 –4
2 5
10-4

51. Solving Quadratic Equations
ALGEBRA 1 LESSON 10-4
56. –1.6
57.
58.
59. (x + 4)(x + 1)
60. (y – 13)(y – 2)
61. (a + 5)(a – 2)
62. (z – 12)(z + 6)
63. (c – 12d)(c – 2d)
64. (t + 2u)(t – u)
 106
 10–5
67. –8.12  100
68. 31,000
,000 5 8 7 9
10-4

52. Solving Quadratic Equations
ALGEBRA 1 LESSON 10-4
1. Solve each equation by graphing the related function. If the equation has no solution, write no solution.
a. 2x2 – 8 = 0
b. x2 + 2 = –2
2. Solve each equation by finding square roots.
a. m2 – 25 = 0
b. 49q2 = 9
3. Find the speed of a 4-kg bowling ball with a kinetic energy of 160 joules. Use the equation E = ms2, where m is the object’s mass in kg, E is its kinetic energy, and s is the speed in meters per second. ±2
no solution ±5 3 7 1 2
about 8.94 m/s
10-4

53. Factoring to Solve Quadratic Equations
ALGEBRA 1 LESSON 10-5
(For help, go to Lessons 2-2 and 9-6.)
Solve and check each equation.
n = – 9 = q + 16 = –3
Factor each expression.
4. 2c2 + 29c p2 + 32p x2 – 21x – 18 a 8
10-5

54. Factoring to Solve Quadratic Equations
ALGEBRA 1 LESSON 10-5
Solutions
n = 2
4n = –4
n = –1
Check: 6 + 4(–1) = 6 + (–4) = 2
2. – 9 = 4
= 13
a = 104
Check: – 9 = 13 – 9 = 4
3. 7q + 16 = –3
7q = –19
q = –2
Check: 7 (–2 ) + 16 = 7(– ) + 16 = – = –3 a 8 a 8
104 8 5 7 5 7 19 7
10-5

55. Factoring to Solve Quadratic Equations
ALGEBRA 1 LESSON 10-5
Solutions (continued)
4. 2c2 + 29c + 14 = (2c + 1)(c + 14)
Check: (2c + 1)(c + 14) = 2c2 + 28c + c + 14 = 2c2 + 29c + 14
5. 3p2 + 32p + 20 = (3p + 2)(p + 10)
Check: (3p + 2)(p + 10) = 3p2 + 30p + 2p + 20 = 3p2 + 32p + 20
6. 4x2 – 21x – 18 = (4x + 3)(x – 6)
Check: (4x + 3)(x – 6) = 4x2 – 24x + 3x – 18 = 4x2 – 21x – 18
10-5

56. Factoring to Solve Quadratic Equations
ALGEBRA 1 LESSON 10-5
Solve (2x + 3)(x – 4) = 0 by using the Zero Product Property.
(2x + 3)(x – 4) = 0
2x + 3 = or x – 4 = 0
Use the Zero-Product Property.
2x = –3
Solve for x.
x = – 3 2 or
x = 4
Check: Substitute – for x. 3 2
Substitute 4 for x.
(2x + 3)(x – 4) = 0
[2(– ) + 3](– – 4) 0
[2(4) + 3](4 – 4) 0
(0)(– 5 ) = 0 1
(11)(0) = 0
10-5

57. Factoring to Solve Quadratic Equations
ALGEBRA 1 LESSON 10-5
Solve x2 + x – 42 = 0 by factoring.
x2 + x – 42 = 0
(x + 7)(x – 6) = 0
Factor using x2 + x – 42
x + 7 = 0 or x – 6 = 0
Use the Zero-Product Property.
x = –7 or x = 6
Solve for x.
10-5

58. Factoring to Solve Quadratic Equations
ALGEBRA 1 LESSON 10-5
Solve 3x2 – 2x = 21 by factoring.
3x2 – 2x = 21
Subtract 21 from each side.
(3x + 7)(x – 3) = 0
Factor 3x2 – 2x – 21.
3x + 7 = 0 or x – 3 = 0
Use the Zero-Product Property
3x = –7
Solve for x.
x = – or x = 3 7 3
10-5

59. Factoring to Solve Quadratic Equations
ALGEBRA 1 LESSON 10-5
The diagram shows a pattern for an open-top box. The total area of the sheet of materials used to make the box is 130 in.2. The height of the box is 1 in. Therefore, 1 in.  1 in. squares are cut from each corner. Find the dimensions of the box.
Define: Let x = width of a side of the box.
Then the width of the material = x = x + 2
The length of the material = x = x + 5
Relate: length  width = area of the sheet
10-5

60. Factoring to Solve Quadratic Equations
ALGEBRA 1 LESSON 10-5
(continued)
Write: (x + 2) (x + 5) = 130
x2 + 7x + 10 = 130
Find the product (x + 2) (x + 5).
x2 + 7x – 120 = 0
Subtract 130 from each side.
(x – 8) (x + 15) = 0
Factor x2 + 7x – 120.
x – 8 = 0 or x + 15 = 0
Use the Zero-Product Property.
x = 8 or x = –15
Solve for x.
The only reasonable solution is 8. So the dimensions of the box are 8 in.  11 in.  1 in.
10-5

61. Factoring to Solve Quadratic Equations
ALGEBRA 1 LESSON 10-5
pages 538–540  Exercises
1. 3, 7
2. –4, 4.5
3. 0, –1
4. 0, 2.5
5. – , –
6. , –
7. 1, –4
8. –2, 7
9. 0, 8
10. 5, 11
11. –2, 5
12. 3, –4
13. –3, –5
14. –4, 7
15. 0, 6
16. 1, 2.5
17. –5, –
18. –2.5, 2.5
, –4
20. 2, 3
, –4
22. 5 cm
23. 5 2 5 3
24. 6 ft  15 ft
25. base: 10 ft
height: 22 ft
26. 2 and 3 or 7 and 8
27. 2q2 + 22q + 60 = 0; –6, –5
28. 6n2 – 5n – 4 = 0; , –
29. 4y2 + 12y + 9 = 0; –
30. a2 + 6a + 9 = 0; –3
31. 2t t + 12 = 0; –1.5, –4
32. x2 – 10x + 24 = 0; 4, 6
33. 2k2 + 11k – 63 = 0; , –9
34. 20y2 + 41y – 9 = 0; , – 2 7 4 5 1 3 4 3 1 2 7 4 8 3 3 2 7 2 1 5 9 4
10-5

62. Factoring to Solve Quadratic Equations
ALGEBRA 1 LESSON 10-5
35. 8 in.  10 in.
36. a. 2 s
b. about 19 ft
37. Answers may vary. Sample:
To solve a quadratic equation,
write the equation in standard
form, factor the quadratic
expression, use the Zero-Product
Property, and solve for the variable.
x2 + 8x = –15
x2 + 8x = 0
(x + 3)(x + 5) = 0
x + 3 = 0 or x + 5 = 0
x = –3  or  x = –5
38. Answers may vary. Sample:
x = 6, a = 2, b = 1;
x = 3, a = 1, b = 11
39. Answers may vary. Sample:
x2 – 2x – 8 = 0
(x – 4)(x + 2) = 0
x – 4 = 0 or x + 2 = 0
x = 4   or   x = –2
40. a. 0, 1; –1, 0
b. 0
41. 0, 4, 6
42. 0, 1, 4
43. 0, 3
44. 0, 7, –10
45. 0, 1, 9
46. 0, 4, –5
47. 4
10-5

63. Factoring to Solve Quadratic Equations
ALGEBRA 1 LESSON 10-5
48. Answers may vary. Samples:
a. x2 – 3x – 40 = 0
b. x2 – x – 6 = 0
c. 2x2 + 19x – 10 = 0
d. 21x2 + x – 10 = 0
49. –1, 1, –5
50. –2, 2, –1
51. D
52. H
53. A
54. C
55. D
56. [2] 3x2 + 20x + 1 = 8
3x2 + 20x – 7 = 0
(3x – 1)(x + 7) = 0
x = , x = –7
[1] appropriate methods,
but with one computational error
57. x2 = 320; 17.9 ft
= r 2; 3.5 ft
59. (2x + 3)(x + 5)
60. (3y – 1)(y – 3)
61. (4t – 3)(t + 2)
62. (3n – 1)(2n + 3)
63. 5a(3a + 2)(a – 4)
64. –2b(3b – 2)(3b – 5) 1 3
10-5

64. Factoring to Solve Quadratic Equations
ALGEBRA 1 LESSON 10-5
1. Solve (2x – 3)(x + 2) = 0.
Solve by factoring.
2. 6 = a2 – 5a 3. 12x + 4 = –9x2 4. 4y2 = 25
–2, 3 2 – 2 3 5 2
–1, 6
10-5

65. Completing the Square Find each square.
ALGEBRA 1 LESSON 10-6
(For help, go to Lessons 9-4 and 9-7.)
Find each square.
1. (d – 4)2 2. (x + 11)2 3. (k – 8)2
Factor.
4. b2 + 10b t2 + 14t n2 – 18n + 81
10-6

66. Completing the Square Solutions
ALGEBRA 1 LESSON 10-6
1. (d – 4)2 = d2 – 2d(4) + 42 = d2 – 8d + 16
2. (x + 11)2 = x2 + 2x(11) = x2 + 22x + 121
3. (k – 8)2 = k2 – 2k(8) + 82 = k2 – 16k + 64
4. b2 + 10b + 25 = b2 + 2b(5) + 52 = (b + 5)2
5. t2 + 14t + 49 = t2 + 2t(7) + 72 = (t + 7)2
6. n2 – 18n + 81 = n2 – 2n(9) + 92 = (n – 9)2
Solutions
10-6

67. Find the value of c to complete the square for x2 – 16x + c.
Completing the Square
ALGEBRA 1 LESSON 10-6
Find the value of c to complete the square for x2 – 16x + c.
The value of b in the expression x2 – 16x + c is –16.
The term to add to x2 – 16x is or 64.
–16 2
10-6

68. Solve the equation x2 + 5 = 50 by completing the square.
ALGEBRA 1 LESSON 10-6
Solve the equation x2 + 5 = 50 by completing the square.
Step 1: Write the left side of x2 + 5 = 50 as a perfect square.
x2 + 5 = 50
x = 50 + 5 2
Add , or , to each side of the equation. 25 4 5 2
x +
200 4 + 25 =
Write x2 + 5x as a square.
Rewrite 50 as a fraction with denominator 4. 5 2
x + =
225 4
10-6

69. Step 2: Solve the equation.
Completing the Square
ALGEBRA 1 LESSON 10-6
(continued)
Step 2: Solve the equation.
Find the square root of each side. 5 2
x + =
225 4 15
Simplify. 5 2
x + = 15 or –
Write as two equations.
x = 5 or x = –10
Solve for x.
10-6

70. Add 16 to each side of the equation.
Completing the Square
ALGEBRA 1 LESSON 10-6
Solve x2 + 10x – 16 = 0 by completing the square. Round to the nearest hundredth.
Step 1: Rewrite the equation in the form x2 + bx = c and complete the square.
x2 + 10x – 16 = 0
x2 + 10x = 16
Add 16 to each side of the equation.
x2 + 10x + 25 =
Add , or 25, to each side of the equation. 10 2
(x + 5)2 = 41
Write x2 + 10x +25 as a square.
10-6

71. Step 2: Solve the equation.
Completing the Square
ALGEBRA 1 LESSON 10-6
(continued)
Step 2: Solve the equation.
x + 5 = ± 41
Find the square root of each side.
Use a calculator to find 41
x + 5
± 6.40
x + 5
6.40 or
–6.40
Write as two equations.
Subtract 5 from each side.
x 6.40 – 5 or x –6.40 – 5
x 1.40 or x –11.40
Simplify
10-6

72. Relate: length  width = area
Completing the Square
ALGEBRA 1 LESSON 10-6
Suppose you wish to section off a soccer field as shown in the diagram to run a variety of practice drills. If the area of the field is 6000 yd2, what is the value of x?
Define: width = x = x + 20
length = x + x = 2x + 20
Relate: length  width = area
Write: (2x + 20)(x + 20) = 6000
2x2 + 60x = 6000
Step 1: Rewrite the equation in the form x2 + bx = c.
2x2 + 60x = 6000
2x2 + 60x = 5600
Subtract 400 from each side.
x2 + 30x = 2800
Divide each side by 2.
10-6

73. Step 2: Complete the square.
Completing the Square
ALGEBRA 1 LESSON 10-6
(continued)
Step 2: Complete the square.
x2 + 30x =
Add , or 225, to each side. 30 2
(x + 15)2 = 3025
Write x2 + 30x as a square.
Step 3: Solve each equation.
(x + 15) = ±
Take the square root of each side.
x + 15 = ± 55
Use a calculator.
x + 15 = 55 or x + 15 = –55
x = 40 or x = –70
Use the positive answer for this problem.
The value of x is 40 yd.
10-6

74. Completing the Square pages 544–546 Exercises 12. 19, –17 24. 7, –2
ALGEBRA 1 LESSON 10-6
pages 544–546  Exercises
1. 49
2. 16
3. 400
4. 9
5. 144
6. 324
7. 4, –12
, –3.06
9. –5, –17
, –7.24
11. 9, –29
12. 19, –17
13. 7, –5
14. –2.17, –7.83
15. 11, 1
, –4.19
, –5.82
18. 22, –31
19. 1
20. 4
21.
, –4.16
23. 5, –1
24. 7, –2
25. a. (2x + 1)(x + 1)
b. 2x2 + 3x + 1 = 28
c. 3
26. –0.27, –3.73
27. –3, –4
28. 4, –10
29. 6, 2
, 1.68
31. no solution
, –1.87
, –0.12
34. –4, –5 81
100
10-6

75. d. No; the answers in part (b) were rounded.
Completing the Square
ALGEBRA 1 LESSON 10-6
35. a. = 50 – 2w
b. w(50 – 2w) = 150; 21.5, 3.5
c. 7 ft  21.5 ft or 43 ft  3.5 ft
d. No; the answers in part
(b) were rounded.
36. The student did not divide
each side of the equation by 4.
37. Answers may vary. Sample:
Add 1 to each side of the
equation, and then complete
the square by adding 225 to
each side of the equation.
Write x2 + 30x as the
square (x + 15)2 and add 1
and 225 to get 226. Then
take square roots and solve
the resulting equations.
38. Answers may vary. Sample:
x2 + 10x – 50 = 0
x2 + 10x = 50
x2 + 10x + 25 =
(x + 5)2 = ± 75
x + 5 = ± 8.7
x ± 8.7
x or x –8.7
x   or  x –13.7
, –1.16
, 1.17
ft by 14.2 ft
42. a. 6x2 + 28x
b. 6x2 + 28x = 384
c. 13 in.  6 in.  6 in.
10-6

76. The value of x is about 13.18 cm. [1] appropriate methods,
Completing the Square
ALGEBRA 1 LESSON 10-6
43. a. A = x2 + 5x + 1
b. 6.86
c ft2
44. a. 3 ± 5
b. (3, –5)
c. Answers may vary.
Sample: p is the
x-coordinate of
the vertex.
45. B
46. I
47. D 7 2
48. [2] (x)(x + x + 4) = 200
(x)(2x + 4) = 200
(x)(x + 2) = 200
x2 + 2x = 200
x2 + 2x + 1 = 201
(x + 1)2 = 201
x ±14.18
x    x –15.18
The value of x is about cm.
[1] appropriate methods,
but with one computational error 1 2 1 2
10-6

77. [3] appropriate methods, but with one computational error
Completing the Square
ALGEBRA 1 LESSON 10-6
53. – ,
54. –
55. – ,
56. (x + 2)2
57. (t – 11)2
58. (b + 5)(b – 5)
59. (4c + 3)2
60. (7s + 13)(7s – 13)
61. 2m(2m + 3)(2m – 3)
62. (5m + 12)2
63. (20k + 3)(20k – 3)
64. (16g – 11)(16g + 11) 8 3 8 3
65. r 12
66. p13
67. –y
68.
69. –
70. t 29
49. [4] a. (8 + x)(12 + x) = 2 • (8 • 12)
x2 + 20x – 96 = 0
b. x2 + 20x = 96
x2 + 20x = 196
(x + 10)2 = 196
x + 10 = ±14
x = 4
c. 12 ft by 16 ft
[3] appropriate methods,
but with one computational error
[2] part (c) not done
[1] no work shown
50. –3, 7
51. –6, –5
52. 0, 5 3 2 1 6 5 2 1
m40 1 w
10-6

78. Completing the Square
ALGEBRA 1 LESSON 10-6
Solve each equation by completing the square. If necessary, round to the nearest hundredth.
1. x2 + 14x = –43
2. 3x2 + 6x – 24 = 0
3. 4x2 + 16x + 8 = 40
–9.45, –4.55
–4, 2
–5.46, 1.46
10-6

79. Using the Quadratic Formula
ALGEBRA 1 LESSON 10-7
(For help, go to Lesson 10-6.)
Find the value of c to complete the square for each expression.
1. x2 + 6x + c 2. x2 + 7x + c 3. x2 – 9x + c
Solve each equation by completing the square.
4. x2 – 10x + 24 = 0 5. x2 + 16x – 36 = 0
6. 3x2 + 12x – 15 = x2 – 2x – 112 = 0
10-7

80. Using the Quadratic Formula
ALGEBRA 1 LESSON 10-7
Solutions
1. x2 + 6x + c; c = = 32 = x2 + 7x + c; c = =
3. x2 – 9x + c; c = =
4. x2 – 10x + 24 = 0; 5. x2 + 16x – 36 = 0;
c = = (–5)2 = c = = 82 = 64
x2 – 10x = – x2 + 16x = 36
x2 – 10x = – x2 + 16x =
(x – 5)2 = (x + 8)2 = 100
(x – 5) = ± (x + 8) = ±10
x – 5 = 1   or x – 5 = –1 x + 8 = 10  or x + 8 = –10
x = 6   or   x = x = 2   or  x = –18 6 2 7 2 49 4 –9 2 81 4
–10 2 16 2
10-7

81. Using the Quadratic Formula
ALGEBRA 1 LESSON 10-7
Solutions (continued)
6. 3x2 + 12x – 15 = x2 – 2x – 112 = 0
3(x2 + 4x – 5) = (x2 – x – 56) = 0
x2 + 4x – 5 = 0; x2 – x – 56 = 0;
c = = 22 = c = =
x2 + 4x = x2 – x = 56
x2 + 4x + 4 = x2 – x = 56 +
(x + 2)2 = (x – )2 =
(x + 2) = ± x – = ±
x + 2 = 3   or x + 2 = –3 x – = or x – =
x = 1   or   x = – x = or x = –7 4 2 –1 2 1 4 1 4 1 4 1 2
225 4 1 2 15 2 1 2 15 2 1 2
–15 2
10-7

82. Using the Quadratic Formula
ALGEBRA 1 LESSON 10-7
Solve x2 + 2 = –3x using the quadratic formula.
x2 + 3x + 2 = 0
Add 3x to each side and write in standard form.
x =
–b ± b2 – 4ac 2a
Use the quadratic formula.
x =
–3 ± (–3)2 – 4(1)(2)
2(1)
Substitute 1 for a, 3 for b, and 2 for c.
x =
–3 ± 1 2
Simplify.
x =
–3 + 1 2
–3 – 1 or
Write two solutions.
x = –1 or x = –2
Simplify.
10-7

83. Using the Quadratic Formula
ALGEBRA 1 LESSON 10-7
(continued)
Check: for x = –1 for x = –2
(–1)2 + 3(–1)
(–2)2 + 3(–2)
1 –
4 –
0 = 0
10-7

84. Using the Quadratic Formula
ALGEBRA 1 LESSON 10-7
Solve 3x2 + 4x – 8 = 0. Round the solutions to the nearest hundredth.
x =
–b ± b2 – 4ac 2a
Use the quadratic formula.
x =
–4 ± – 4(3)(–8)
2(3)
Substitute 3 for a, 4 for b, and –8 for c.
–4 ± 6
x = or
Write two solutions. – 6
–4 –
Use a calculator. x – 6
–4 – or x
x –2.43
Round to the nearest hundredth.
10-7

85. Using the Quadratic Formula
ALGEBRA 1 LESSON 10-7
A child throws a ball upward with an initial upward velocity of 15 ft/s from a height of 2 ft. If no one catches the ball, after how many seconds will it land? Use the vertical motion formula h = –16t2 + vt + c, where h = 0, v = velocity, c = starting height, and t = time to land. Round to the nearest hundredth of a second.
Step 1:  Use the vertical motion formula.
h = –16t2 + vt + c
0 = –16t2 + 15t + 2  Substitute 0 for h, 15 for v, and 2 for c.
Step 2: Use the quadratic formula.
x =
–b ± b2 – 4ac 2a
10-7

86. Using the Quadratic Formula
ALGEBRA 1 LESSON 10-7
(continued)
t =
–15 ± – 4(–16)(2)
2(–16)
Substitute –16 for a, 15 for b, 2 for c, and t for x.
–15 ±
–32
t =
Simplify.
–15 ±
t = –
–32 or
–15 – 18.79
Write two solutions. t
–0.12 or
1.06
Simplify. Use the positive answer because it is the only reasonable answer in this situation.
The ball will land in about 1.06 seconds.
10-7

87. Using the Quadratic Formula
ALGEBRA 1 LESSON 10-7
Which method(s) would you choose to solve each equation? Justify your reasoning.
a. 5x2 + 8x – 14 = 0
Quadratic formula; the equation cannot be factored easily.
b. 25x2 – 169 = 0
Square roots; there is no x term.
c. x2 – 2x – 3 = 0
Factoring; the equation is easily factorable.
d. x2 – 5x + 3 = 0
Quadratic formula, completing the square, or graphing; the x2 term is 1, but the equation is not factorable.
e. 16x2 – 96x = 0
Quadratic formula; the equation cannot be factored easily and the numbers are large.
10-7

88. Using the Quadratic Formula
ALGEBRA 1 LESSON 10-7
pages 550–552  Exercises
1. –1, –1.5
2. 2.8, –6
3. 1.5
4. –0.67, –15
, –0.25
6. –4, –9
, –16
8. 13, –8.5
9. 16, –2.4
, –2.67
, 1.58
, –0.77
, –3.03
, –0.17
16. a. 0 = –16t t + 3
b. t ; 0.8 s
17. a. 0 = –16t t + 3.5
b. t ; 3.2 s
18. Completing the square or graphing;
the x2 term is 1 but the equation is not factorable.
19. Factoring or square roots; the equation
is easily factorable and there is no x term.
20. Quadratic formula; the equation
cannot be factored.
21. Quadratic formula; the equation
22. Factoring; the equation is easily factorable.
10-7

89. Using the Quadratic Formula
ALGEBRA 1 LESSON 10-7
34. About 2.1s
35. Answers may vary. Sample: You solve
the linear equation using transformations
and you solve the quadratic equation
using the quadratic formula.
ft and 5.40 ft
cm and 7.44 cm
38. Answers may vary. Sample: A rectangle
has length x. Its width is 5 feet longer
than three times the length. Find the
dimensions if its area is 182 ft2.
7 ft  26 ft
23. Quadratic formula;
the equation cannot be factored.
24. 6, –6
, –1.54
, –1.41
, –2.61
28. 2
29. 3, –3
, –0.39
, –1
, –1.43
33. a. 7 ft  8 ft
b. x(x + 1) = 60, 7.26 ft  8.26 ft
10-7

90. Using the Quadratic Formula
ALGEBRA 1 LESSON 10-7
53. (6v – 5)(2v + 7)
54. (2x – 1)(3x – 5)
55. (5t + 3)(3t + 2)
39. if the expression b2 – 4ac
equals zero
40. about 1.9 s
41. a. Check students’ work.
b million
c. 2007
42. a. s = –
b. 6.5
46. [2]
–1.8, 3.7
[1] correct substitution into
quadratic formula,
with one calculation error
, 8.46
48. –1, –2
, –6.1
50. (2c + 5)(c + 3)
51. (3z – 2)(z + 4)
52. (5n + 2)(n – 7)
11 ± (112) – 4(6)(–40)
2(6)
x = a b
43. a  107 lb
b. 3.3  104 tons
c s
44. D
45. F
10-7

91. Using the Quadratic Formula
ALGEBRA 1 LESSON 10-7
1. Solve 2x2 – 11x + 12 = 0 by using the quadratic formula.
2. Solve 4x2 – 12x = 64. Round the solutions to the nearest hundredth.
3. Suppose a model rocket is launched from a platform 2 ft above the ground with an initial upward velocity of 100 ft/s. After how many seconds will the rocket hit the ground? Round the solution to the nearest hundredth.
1.5, 4
–2.77, 5.77
6.27 seconds
10-7

92. Using the Discriminant
ALGEBRA 1 LESSON 10-8
(For help, go to Lessons 1-6 and 10-7.)
Evaluate b2 – 4ac for the given values of a, b, and c.
1. a = 3, b = 4, c = 8 2. a = –2, b = 0, c = 9
3. a = 11, b = –5, c = 7
Solve using the quadratic formula. If necessary, round to the
nearest hundredth.
4. 3x2 – 7x + 1 = x2 + x – 1 = 0 6. x2 – 12x + 35 = 0
10-8

93. Using the Discriminant
ALGEBRA 1 LESSON 10-8
Solutions
1. b2 – 4ac for a = 3, b = 4, c = 8: 42 – 4(3)(8) = 16 – 96 = –80
2. b2 – 4ac for a = –2, b = 0, c = 9: 02 – 4(–2)(9) = = 72
3. b2 – 4ac for a = 11, b = –5, c = 7: (–5)2 – 4(11)(7) = 25 – 308 = –283
4. 3x2 – 7x + 1 = 0; x = for a = 3, b = –7, c = 1:
x = = =
x =   or  x =
–b ± b2 – 4ac 2a
–(–7) ± (–7)2 – 4(3)(1)
2(3)
7 ± – 12 6
7 ± 6 6
7 – 6
10-8

94. Using the Discriminant
ALGEBRA 1 LESSON 10-8
Solutions (continued)
5. 4x2 + x – 1 = 0; x = for a = 4, b = 1, c = –1:
x = = =
x =   or  x = –0.64
–b ± b2 – 4ac 2a
–1 ± – 4(4)(–1)
2(4)
–1 ± 8
–1 ± 8 – 8
–1 – 8
6. x2 – 12x + 35 = 0; x = for a = 1, b = –12, c = 35:
x = =
x = =
x = = = 7  or  x = = = 5
–b ± b2 – 4ac 2a
–(–12) ± (–12)2 – 4(1)(35)
2(1)
12 ± – 140 2
12 ± 4 2
12 ± 2 2
12 + 2 2 14 2
12 – 2 2 10 2
10-8

95. Using the Discriminant
ALGEBRA 1 LESSON 10-8
Find the number of solutions of x2 = –3x – 7 using the discriminant.
x2 + 3x + 7 = 0 Write in standard form.
b2 – 4ac = 32 – 4(1)(7) Evaluate the discriminant. Substitute for a, b, and c.
= 9 – 28 Use the order of operations.
= –19 Simplify.
Since –19 < 0, the equation has no solution.
10-8

96. Using the Discriminant
ALGEBRA 1 LESSON 10-8
A football is kicked from a starting height of 3 ft with an initial upward velocity of 40 ft/s. Will the football ever reach a height of 30 ft?
h = –16t2 + vt + c Use the vertical motion formula.
30 = –16t2 + 40t + 3 Substitute 30 for h, 40 for v, and 3 for c.
0 = –16t2 + 40t – 27 Write in standard form.
b2 – 4ac = (40)2 – 4 (–16)(–27) Evaluate the discriminant.
= 1600 – 1728 Use the order of operations.
= –128 Simplify.
The discriminant is negative. The football will never reach a height of 30 ft.
10-8

97. Using the Discriminant
ALGEBRA 1 LESSON 10-8
21. 2
22. 2
23. 0
24. 2
25. a. S = –0.75p2 + 54p
b. no
c. $36
d. If a product is too expensive,
fewer people will buy it.
26. a. k > 4
b. k = 4
c. k < 4
27. a. A2 ^ 2 – 4;
A2 ^ 2 – 8
b. |b| < 2
pages 556–558  Exercises
1. A
2. C
3. B
4. 0
5. 1
6. 2
7. 2
8. 2
9. 2
10. 0
11. 2
12. 2
13. 1
14. 2
15. 0
16. none
17. No; the discriminant
is negative.
18. a. yes
b. no
c. no
d. no
19. 0
20. 0
10-8

98. Using the Discriminant
ALGEBRA 1 LESSON 10-8
38. never
39. sometimes
40. always
41. 2; since the parabola crosses the x-axis
once, it must cross again.
42. y = 2x2 + 8x + 10 has a vertex closer to
the x-axis; its discriminant is closer
to zero.
43. C
44. G
45. B
46. D
47. A
30. a. 16; 5, 1
b. 81; 4, –5
c. 73; 3.89, –0.39
d. Rational; the square root
of a discriminant that is a
perfect square is a pos. integer.
31. no
32. no
33. yes; 1, –1.25
34. yes; –1,
35. no
36. yes; 2.5, –1
37. Answers may vary. Sample:
Use values for a, b, and c such
that the discriminant is positive. 2 3
10-8

99. Using the Discriminant
ALGEBRA 1 LESSON 10-8
55. $
56. $312.88
57. $
58. arithmetic
59. arithmetic
60. geometric
61. arithmetic
48. [2] x(25 – x) = 136
x2 – 25x = 0
x =
= = 17 or 8
yes, 17 cm by 8 cm
[1] no work shown OR appropriate
methods with one computational error
, –1.5
, –3.83
, –0.27
, –2.79
, 0.13
54. 2, 1.5
–(–25) ± (–25)2 – 4(1)(136)
2(1)
25 ± 9 2
10-8

100. Using the Discriminant
ALGEBRA 1 LESSON 10-8
Find the number of solutions for each equation.
1. 3x2 – 4x = 7
2. 4x2 = 4x – 1
3. –3x2 + 2x – 12 = 0
4. A ball is thrown from a starting height of 4 ft with an initial upward velocity of 30 ft/s. Is it possible for the ball to reach a height of 18 ft?
two
one
none
yes
10-8

101. Choosing a Model Graph each function. 1. y = 3x – 1 2. y = x + 2
ALGEBRA 1 LESSON 10-9
(For help, go to Lessons 6-2, 8-7 and 10-1.)
Graph each function.
1. y = 3x – 1 2. y = x + 2
3. y = 2x 4. y = x
5. y = x y = 2x2 – 1 1 4 3
10-9

102. Choosing a Model Solutions 1. y = 3x – 1 2. y = x + 2
ALGEBRA 1 LESSON 10-9
Solutions
1. y = 3x – 1 2. y = x + 2
3. y = 2x 4. y = x
5. y = x y = 2x2 – 1 1 4 1 3
10-9

103. Choosing a Model
ALGEBRA 1 LESSON 10-9
Graph each set of points. Which model is most appropriate for each set?
a. (–2, 2.25), (0, 3),
(1, 4) (2, 6)
b. (–2, –2), (0, 2),
(1, 4), (2, 6)
c. (–1, 5), (2, 11), (0, 3),
(1, 5), (–2, 11)
exponential model
linear model
quadratic model
10-9

104. There is a common second difference, 2.8.
Choosing a Model
ALGEBRA 1 LESSON 10-9
a. Which kind of function best models the data below? Write an equation to model the data.
Step 1: Graph the data
Step 2: The data appear to be quadratic. Test for a common second difference.
x y
0 0
1 1.4
2 5.6
3 12.6
4 22.4 +1
+ 1.4
+ 4.2
+ 7.0
+ 9.8
+ 2.8
x y
0 0
1 1.4
2 5.6
3 12.6
4 22.4
There is a common second difference, 2.8.
10-9

105. Step 3: Write a quadratic model. y = ax2
Choosing a Model
ALGEBRA 1 LESSON 10-9
a. (continued)
Step 3: Write a quadratic model.
y = ax2
5.6 = a(2)2 Use a point other than(0, 0) to find a.
5.6 = 4a Simplify.
1.4 = a Divide each side by 4.
y = 1.4x2 Write a quadratic function.
Step 4: Test two points other than (2, 5.6) and (0, 0)
y = 1.4(1)2 y = 1.4(3)2
y = 1.4 • 1 y = 1.4 • 9
y = 1.4 y = 12.6
(1, 1.4) and (3, 12.6) are both data points.
The equation y = 1.4x2 models the data.
10-9

106. Choosing a Model
ALGEBRA 1 LESSON 10-9
b. Which kind of function best models the data below? Write an equation to model the data.
Step 1: Graph the data
Step 2: The data appear to suggest an exponential model. Test for a common ratio.
x y
–1 4
0 2
1 1
2 0.5
3 0.25 +1
2  4 = 0.5
1  2 = 0.5
0.5  1 = 0.5
0.25  0.5 = 0.5
x y
–1 4
0 2
1 1
2 0.5
3 0.25
10-9

107. Step 3: Write an exponential model.
Choosing a Model
ALGEBRA 1 LESSON 10-9
b. (continued)
Step 3: Write an exponential model.
Relate: y = a • bx
Define: Let a = the initial value, 2.
Let b = the decay factor, 0.5.
Write: y = 2 • 0.5x
Step 4: Test two points other than (0, 2).
y = 2 • 0.51 y = 2 • 0.52
y = 2 • 0.5 y = 2 • 0.25
y = 1.0 y = 0.5
(1, 1) and (2, 0.5) are both data points.
The equation y = 2 • 0.5x models the data.
10-9

108. Step 1: Graph the data to decide which model is most appropriate.
Choosing a Model
ALGEBRA 1 LESSON 10-9
Suppose you are studying deer that live in an area. The data in the table was collected by a local conservation organization. It indicates the number of deer estimated to be living in the area over a five-year period. Determine which kind of function best models the data. Write an equation to model the data.
Year Estimated
Population
0 90
1 69
2 52
3 40
4 31
Step 1: Graph the data to decide which model is most appropriate.
The graph curves, and it does not look quadratic. It may be exponential.
10-9

109. Step 2: Test for a common ratio.
Choosing a Model
ALGEBRA 1 LESSON 10-9
(continued)
Step 2: Test for a common ratio.
Year Estimated
Population
0 90
1 69
2 52
3 40
4 31 +1
69 
52 
40 
31 
The common ratio is roughly 0.77.
The population of deer is roughly 0.77 times its value the previous year.
10-9

110. Step 3: Write an exponential model.
Choosing a Model
ALGEBRA 1 LESSON 10-9
(continued)
Step 3: Write an exponential model.
Relate: y = a • bx
Define: Let a = the initial value, 90.
Let b = the decay factor, 0.77.
Write: y = 90 • 0.77x
Step 4:  Test two points other than (0, 90).
y = 90 • y = 90 • 0.772
y y 53
The predicted value (1, 69) matches the corresponding data point. The point (2, 53) is close to the data point (2, 52).
The equation y = 90 • 0.77x models the data.
10-9

111. Choosing a Model pages 563–566 Exercises 1. quadratic 2. linear 3.
ALGEBRA 1 LESSON 10-9
pages 563–566  Exercises 1.
quadratic 2.
linear 3.
exponential 4.
quadratic 5.
exponential 6.
linear
7. quadratic; y = 1.5x2
8. linear; y = 2x – 5
9. quadratic; y = 2.8x2
10-9

112. Choosing a Model 10. exponential; y = 1 • 1.2x 14. a. exponential
ALGEBRA 1 LESSON 10-9
10. exponential; y = 1 • 1.2x
11. exponential; y = 5 • 0.4x
12. linear; y = – x + 2
13. a.
linear
b. 65, 64, 64; yes
c. 64
d. y = 64x – 5
14. a. exponential
b. y = 16,500 • 0.88x
15. a. 41, 123, 206
b. 82, 83
c. d = 41t 2
d cm
16. a.
linear
b. 5 years
c. 600, 600, 600; 120, 120, 120
d. p = 120t 1 2
10-9

113. 18. Answers may vary. Sample: Linear data have a common
Choosing a Model
ALGEBRA 1 LESSON 10-9
22. y = –0.336x2 – 0.219x
23. y = –1.1x + 3.5
24. y = • 2.582x
25. a. i.
ii.
17. a. 5
b. 398, 429, 407, 389;
79.6, 85.8, 81.4, 77.8
c. 81.2
d. p = 81.2t
e million,
or about 6.9 billion
18. Answers may vary. Sample:
Linear data have a common
first difference, quadratic data
have a common second
difference, and exponential
data have a common ratio.
19. y = 0.875x2 – 0.435x
20. y = • 0.770x
21. y = 2.125x2 – 4.145x
x y
1 –2
2   1
3   6
4 13
5 22
) 3 ) 2
) 5 ) 2
) 7 ) 2
) 9
x y
1  3
2 12
3 27
4 48
5 75
)  9 ) 6
) 15 ) 6
) 21 ) 6
) 27
10-9

114. c. When second differences are the same, the data are
Choosing a Model
ALGEBRA 1 LESSON 10-9
26. Answers may vary.
Sample:
27. a. quadratic
b. d = 13.6t 2
c ft
28. Check students’ work.
iii.
b. The second common
difference is twice
the coefficient of x2.
c. When second differences
are the same, the data are
quadratic. You can determine
the coefficient of x2 by
dividing the second difference by 2.
x y
1 –1
2   6
3 21
4 44
5 75
)  7 ) 8
) 15 ) 8
) 23 ) 8
) 31
x y
0  5
2 13
4 29
6 53
10-9

115. inaccurate evaluation 33. [4] a. linear b. d = –2.5n + 43.5 c. 18
Choosing a Model
ALGEBRA 1 LESSON 10-9
30. B
31. H
32. [2] p = 33,500(1.014)n,
33,500(1.014) ,497
[1] correct formula,
inaccurate evaluation
33. [4] a. linear
b. d = –2.5n
c. 18
[3] appropriate methods,
but with one computational error
[2] part (c) not answered
[1] no work shown
34. 1
35. 0
36. 2
29. a. 1.85, 1.28, 1.45, 1.43
b. 139, 85, 174, 240
c. –54, 89, 66
d. 1.85; the ratio is much
larger than the other ratios.
e. 85; the difference is much
smaller than the other differences. f.
10-9

116. Choosing a Model
ALGEBRA 1 LESSON 10-9
44. –32
45.
47. 16
37.
38.
39.
40.
41.
42. 2 27
10-9

117. Choosing a Model
ALGEBRA 1 LESSON 10-9
Which kind of function best models the data in each table? Write an equation to model the data.
x y
–1 15
0 3
1 0.6
2 0.12 1.
x y
–1 –5
0 –3
1 –1
2 1
3 3 2.
x y
–1 2.2
0 0
1 2.2
2 8.8
3 19.8 3.
exponential;
y = 3 • 0.2x
linear;
y = 2x – 3
quadratic;
y = 2.2x2
10-9

118. Quadratic Equations and Functions
ALGEBRA 1 CHAPTER 10
1. D
2. C
3. A
4. B
5. x = 0, (0, –7); min.
6. x = 1.5, (1.5, –0.25); min.
7. x = 2.5, (2.5, 11.5); max.
8. x = –6, (–6, –18); min. 9.
10.
11.
12.
13.
14.
15. Answers may vary. Sample:
You can tell how wide it is
and whether it opens
upward or downward.
10-A

119. Quadratic Equations and Functions
ALGEBRA 1 CHAPTER 10
16. 1
17. 0
18. 2
19. 2
20.
21.
23. 40
24.
26. 5, 6
27. 11, 12
28. 18, 19
29. –3, –2
30. 1
31. no solution
32. 2
33. 2 2 3
34. 1
35. 5, –5
, –1
37. 2, –8
38. 1, –2
, –4.24
40. 1, –2.33
41. Answers may vary.
Sample: y = –x2 + 4;
= r 2,
2.1 ft
= 2w 2,
w = 20 ft,
= 40 ft
44. quadratic;
y = x2
45. exponential;
y = (2x)
46. linear;
y = x + 2
47. exponential;
y = 40(0.5x) 1 2 1 2
10-A