Answers to Algebra 2 Unit 2 Practicepshs.psd202.org/documents/bmccormi/1508950153.pdfSpringBoard...

A1© 2015 College Board. All rights reserved. SpringBoard Algebra 2, Unit 2 Practice

Answers to Algebra 2 Unit 2 PracticeLeSSon 7-1 1. a. A(l ) 5 40l 2 l 2

b. The Area of a Rectangle with Perimeter 80

10 20 30 40 50l

A(l)

100

200

300

400

500

Area

(cm

2 )

Length (cm)

c. Yes; the length of a rectangle that has an area of 256 cm2 is 32 cm. The width is 8 cm. Both 8 and 32 satisfy the equation, 256 5 40l 2 l2.

d. 10 cm 3 30 cm or 30 cm 3 10 cm

e. domain: 0 , x , 40, (0, 40), {x | x ∈ , 0 , x , 40};

range: 0 , A # 400, (1, 400), {A | A ∈ , 0 , A # 400}

f. 400 cm2; the graph shows the maximum to be at 400. This occurs when the length is 20 cm. The width will also be 20 cm since the perimeter will be 80 cm when both the length and width are 20 cm. This rectangle is a square.

2. C

3. 6 horses; with 240 ft of fencing, the maximum area of the corral would be 3600 sq ft. This is enough space for 6 horses.

4. 400 sq ft; A(20) 5 40(20) 2 202 5 400

5. The maximum value is represented by the y-coordinate of the highest point on the graph.

LeSSon 7-2 6.

x2 29x

23x 27

7. a. (x 2 7)(x 1 5)

b. (x 2 9)(x 2 6)

c. (x 2 2)(x 2 2) or (x 2 2)2

d. (x 2 11)(x 1 11)

e. (2x 2 5)(x 1 3)

f. (3x 2 8)(2x 1 3)

g. (3x 2 4)(x 2 2)

h. (5x 2 3)(2x 1 3)

i. (3x 1 4)(4x 1 3)

j. (5x 1 3)(3x 2 2)

8. D

9. Since c is negative, the constant terms in the factored form are different. Since b is positive, the constant term with the greater absolute value will be positive.

10. No; there are no factors of 2 whose sum is 2.

LeSSon 7-3

11. a. x 5 12

, x 5 232

b. x 5 253

, x 5 2

c. x 5 23

, x 5 2

d. x 5 212

, x 5 52

e. x 5 232

, x 5 23

f. x 5 35

, x 521

g. x 5 234

, x 5212

h. x 5 56

, x 51

i. x 5 67

, x 523

j. x 5 245

, x 5 32


12. a. x2 2 x 2 6 5 0

b. x2 1 5x 1 4 5 0

c. x2 2 2x 2 15 5 0

d. x2 2 4x 2 21 5 0

e. 2x2 2 x 2 1 5 0

f. 6x2 2 7x 1 2 5 0

g. 12x2 2 x 2 6 5 0

h. 15x2 2 14x 1 3 5 0

13. A

14. To solve a quadratic equation by factoring, you use the Zero Product Property. First, set the equation equal to 0. Next, factor the equation. Since the equation is equal to 0, one or both of the factors must be equal to 0. Set each factor equal to 0 and solve.

15. a.

Garden

12 ft

10 ft

x ft

x ft

b. 4x2 1 44x 2 120 5 360

c. 4(x 1 15)(x 2 4) 5 0

d. The solution x 5 4 shows that the width of the path is 4 ft. The solution x 5 215 should be discarded because a negative measure for the path does not make sense.

LeSSon 7-4 16. x . 1 or x , 25; product is positive when the

value of x results in both factors having the same sign.

17. a. x # 232

or x $ 4

2628210 24 22 0 2 4 6 8 10

b. 24 , x , 2

2628210 24 22 0 2 4 6 8 10

18. C

19. a. 25 , x , 3

b. x , 12

or x . 4

c. x , 20.5 or x . 2.5

d. 23 # x # 3

e. x , 21 or x . 5

f. 28 , x , 1

g. x , 223

or x . 5

h. 24 # x # 4

20. a. l (75 2 l2

) $ 2500

b. l2 2 150l 1 5000 # 0

c. (l 2 100)(l 2 50) # 0

d. 50 # l # 100, 25 # w # 50

LeSSon 8-1 21. a. 6i

b. 11i

c. i 5

d. 2i 6

e. 3i 3

f. 7i 2

g. 4i 3

h. 30i

22. A

23. imaginary axis

real axis22242628210

210

28

26

24

22

2

4

6

8

10

2 4 6 8 10

22 2 2i

4 1 3i

24i

32 i


24. a. 6 1 2i

b. 4 2 4i

c. 25

d. 23 2 6i

e. 7i

25. a. 12 2 x; x(12 2 x) 5 40

b. x 5 6 1 2i, x 5 6 2 2i

LeSSon 8-2 26. a. 12 1 2i

b. 21 1 4i

c. 5 1 5i

d. 24 1 10i

e. 21 2 2i

f. 22 11i

g. 8i

h. 3

27. a. 16 1 11i

b. 19 1 2i

c. 29

d. 236 1 26i

e. 10 2 30i

f. 26 1 7i

g. 7 2 17i

h. 22 2 7i

28. a. 2 i12

12

b. i110

45

1

c. i45

35

1

d. 1 i225

95

e. 2 1 i32

52

f. i52

32

2 2

g. 2 i1013

1513

h. i75

45

1

29. Answers will vary; accept any complex number with an imaginary part of 22i. Sample answer: 2 2 2i; (5 2 2i) 2 (2 2 2i) 5 3 1 0i 5 3.

30. C

LeSSon 8-3 31. a. (3x 1 6i)(3x 2 6i) or 3(x 1 2i)(x 2 2i)

b. (5x 1 7i)(5x 2 7i)

c. 3(x 1 3yi)(x 2 3yi)

d. (x 5 1 10yi)(x 5 2 10yi)

32. a. x 5 25i, x 5 5i

b. x 5 212

i, x 5 12

i

c. x 5 2i 5, x 5 i 5

d. x 5 2 i3 22

, x 5 i3 22

33. D

34. a. x 5 27i, x 5 7i

b. x 5 245

i, x 5 45

i

c. x 5 283

i, x 5 83

i

d. x 5 2910

i, x 5 910

i

35. x 5 2i4 23

, x 5 i4 23 ;

9x2 1 32 5 (3x)2 1 (4 2 )2

5 (3x 1 (4i 2)(3x 2 (4i 2) 5 0 3x 1 4i 2 5 0, 3x 2 4i 2 5 0 3x 5 24i 2 , 3x 5 4i 2

x 5 2i4 23

, x 5 i4 23


LeSSon 9-1

36. x 5 2 6 213

; sample explanation:

Step 1: Add 7 to both sides. 3(x 2 2)2 5 7

Step 2: Divide both sides by 3. (x 2 2)2 5 73

Step 3: Take the square root of x 2 2 5 673

both sides.

Step 4: Rationalize the denominator. x 2 2 5 6213

Step 5: Add 2 to both sides. x 5 2 6213

37. a. x 5 654

b. x 5 653

c. x 5 63 24

d. x 5 6i7 33

e. x 5 232

, x 5 152

f. x 5 29 6 i4 55

g. x 5 4 6 303

h. x 5 22 6 i3 77

38. C

39. a. x2 1 8x 1 16; (x 1 4)2

b. x2 2 14x 1 49; (x 2 7)2

40. a. x 5 24 6 17

b. x 5 3 6 5 33

c. x 5 5 6 22

d. x 5 3 6 i 62

LeSSon 9-2

41. x 5b b ac

a4

2

22 6 2

42. a. x 5 21, x 5 12

b. x 5 1 6 3

c. x 5 212

6 2

d. x 5 12

, x 5 2

e. x 5 23, x 5 23

f. x 5 i5 1520

6

g. x 5 11 5 13

66

h. x 5 13 2 2052

2 6

43. D

44 . a. x 5 23 6 3 2 ; completing the square; the variable terms are already isolated on one side, the coefficient of the x2-term is 1, and the coefficient of the x-term is even. This makes completing the square easy.

b. x 5 23, x 5 5; factoring; the left side of the equation can easily be factored as (x 1 3)(x 2 5).

c. x 5 9 16110

2 6 ; Quadratic Formula; the

coefficient of the x2-term is not 1. This makes the other methods of solving a quadratic equation difficult.

d. x 5 24, x 5 14; taking the square root of both sides. When you add 81 to both sides of the equation, each side is a perfect square.

45. a. t 5 17 65

326

b. about 0.28 s and 0.78 s after the ball is thrown


c. Sample answer: Substitute the times from part b into the original equation to check that they make the left side of the equation approximately equal to 10.

216(0.28)2 1 17(0.28) 1 6.5 ≈ 10

21.2544 1 4.26 1 6.5 ≈ 10

9.5056 ≈ 10•

216(0.78)2 1 17(0.78) 1 6.5 ≈ 10

29.7344 1 13.26 1 6.5 ≈ 10

10.0256 ≈ 10•

LeSSon 9-3 46. b2 2 4ac; the value of the discriminant reveals the

number and nature of the roots: one real rational root (discriminant is zero), two real rational roots (discriminant is positive and a perfect square), two real irrational roots (discriminant is positive but not a perfect square), two complex conjugate roots (discriminant is negative).

47. a. 37; since b2 2 4ac is positive and not a perfect square, there are two real, irrational roots.

b. 247; since b2 2 4ac is negative, there are two complex conjugate roots.

c. 16; since b2 2 4ac is positive and a perfect square, there are two real, rational roots.

d. 0; since b2 2 4ac is zero, there is one real, rational root.

48. B

49. a. 0

b. sample answer: (x 2 1)2

50. a. The radicand will be positive when b2 . 4ac.

b. If the radicand is positive, there are two real solutions.

c. When the radicand is positive and a perfect square, the solution will be rational.

LeSSon 10-1 51. A 52. a. x 5 22; the directrix is horizontal, so the axis

of symmetry is a vertical line through the focus. The focus has an x-coordinate of 22, so the axis of symmetry is the line x 5 22.

b. (22, 2); the vertex is the midpoint of the segment that connects the focus and the directrix. The endpoints of this segment have coordinates (22, 3) and (22, 1), so the vertex has coordinates (22, 2).

c. Opens up; the axis of symmetry is vertical and the focus is above the directrix, so the parabola opens up.

53. a. x 5 112

(y 1 3)2 1 5

b. y 5 x

14( 2) 321 1

c. y 5 2116

x2

d. x 5 218

(y 2 5)2 2 1

54. vertex: (23, 21); axis of symmetry: x 5 23; focus: (23, 0); directrix: y 5 22

55.

x

y

22242628210

210

28

26

24

22

2

4

6

8

10

2 4 6 8 10

LeSSon 10-2 56. a. y 5 x2 2 4x 1 7

b. y 5 x2 1 2x 2 1

c. y 5 x2 2 2x 1 3

d. y 5 x2 1 4x 1 1

57. f (x) 5 x2 1 3x 2 5

58. B


59.

x

y

2428212216220

220

216

212

28

24

4

8

12

16

20

4 8 12 16 20

60. a. (0, 22); sample explanation: For a quadratic function, the axis of symmetry is a vertical line that passes through the vertex, so the axis of symmetry is x 5 1. The point (2, 22) is 1 unit to the right of the axis of symmetry. The point one unit to the left of the axis of symmetry with the same y-coordinate as (2, 22) is (0, 22).

b. y 5 x2 2 2x 2 2

LeSSon 10-3 61. a. A quadratic model is a better fit. Sample

justification: A graph shows the data points are closer to the quadratic model. The values of y first increase and then begin to decrease as x increases, which indicates the shape of a quadratic model.

b. A linear model is a better fit. Sample justification: The values of y increase as x increases without ever decreasing, which indicates the shape of a linear model.

62. a. Rocket A: 22.1x2 1 54.3x 2 3.9

Rocket B: 26.5x2 1 101.6x 2 9.9

b. Rocket B. Sample explanation: Graph both quadratic models on the same coordinate grid. The graphs show that Rocket B reaches a greater height than Rocket A.

c. Rocket B will hit the ground around 10 seconds sooner than Rocket A. Sample explanation: I set the height y of each quadratic model equal to 0 and used the Quadratic Formula to solve for the time x. The solutions show that Rocket B will hit the ground after about 15 seconds and Rocket A will hit the ground after about 25 seconds, or about 10 seconds later.

63. A

64. Three. Three noncollinear points determine a parabola, so you can perform a quadratic regression if you have at least 3 data points.

65. a. y 5 20.11x2 1 54.74x 1 3878.57

b. Yes; the monthly revenue increases and then decreases as the selling price increases, which indicates a quadratic model could be a good fit for the data.

c. Answers will vary but should be close to $250.

LeSSon 11-1 66. a. translated 3 units down

x

f(x)

22242628210

210

28

26

24

22

2

4

6

8

10

2 4 6 8 10


b. translated 3 units to the right

x

f(x)

22242628210

210

28

26

24

22

2

4

6

8

10

2 4 6 8 10

c. translated 2 units left and 3 units up

x

f(x)

22242628210

210

28

26

24

22

2

4

6

8

10

2 4 6 8 10

d. translated 2 units right and 3 units down

x

f(x)

22242628210

210

28

26

24

22

2

4

6

8

10

2 4 6 8 10

67. a. translated 4 units up; g(x) 5 x2 1 4

b. translated 3 units left; h(x) 5 (x 1 3)2

c. translated 1 unit right and 5 units down; j(x) 5 (x 2 1)2 25

d. translated 4 units left and 3 units down; k(x) 5 (x 1 4)2 23

68. B

69. (22, 24). Sample explanation: The graph of g(x) is the graph of f(x) 5 x2 translated 2 units to the left and 4 units down. The vertex of f(x) is (0, 0), so the vertex of g(x) will be 2 units to the left and 4 units down from (0, 0) at (22, 24).

70. x 5 3. Sample explanation: The graph of h(x) is the graph of f(x) 5 x2 translated 3 units to the right and 1 unit up. The axis of symmetry of f(x) is x 5 0, so the axis of symmetry of h(x) will be 3 units to the right of x 5 0 at x 5 3.


LeSSon 11-2 71. a. a translation of 3 units up and a reflection over

the x-axis

x

f(x)

22242628210

210

28

26

24

22

2

4

6

8

10

2 4 6 8 10

b. a vertical stretch by a factor of 2 and a translation down 1 unit

x

f(x)

22242628210

210

28

26

24

22

2

4

6

8

10

2 4 6 8 10

c. a horizontal stretch by a factor of 2 and a translation down 3 units

x

f(x)

22242628210

210

28

26

24

22

2

4

6

8

10

2 4 6 8 10

d. a vertical shrink by a factor of 12

, a translation

up 1 unit, and a reflection over the x-axis

x

f(x)

22242628210

210

28

26

24

22

2

4

6

8

10

2 4 6 8 10

72. a. translated 3 units right, stretched vertically by a factor of 2, and reflected over the x-axis, g(x) 5 22(x 2 3)2

b. translated 1 unit left and vertically shrunk by a

factor of 13

, h(x) 5 13

(x 1 1)2

c. horizontally stretched by a factor of 2 and translated up 1 unit, j(x) 5 (1

2x)2 1 1

d. translated down 5 units and vertically stretched by a factor of 3, k(x) 5 3x2 2 5


73. Sample answer: The graph of g(x) 5 12

x2 is a

vertical shrink of f (x) by a factor of 12

. The graph of

h(x) 5 (12 x)2

is a horizontal stretch of f (x) by a

factor of 2.

74. C

75. (21, 2). Sample explanation: The graph of h(x) is a translation 1 unit to the left of the graph of f(x) 5 x2 . This translation moves the vertex from (0, 0) to (21, 0). The translation is followed by a vertical stretch by a factor of 3. This stretch does not change the position of the vertex. The stretch is then followed by a translation 2 units up. This moves the vertex from (21, 0) to (21, 2).

LeSSon 11-3 76. a. g(x) 5 (x 2 3)2 1 5; translated 3 units right and

5 units up

x

g(x)

22242628210

210

28

26

24

22

2

4

6

8

10

2 4 6 8 10

b. h(x) 5 (x 1 5)2 2 1; translated 5 units left and 1 unit down

x

h(x)

22242628210

210

28

26

24

22

2

4

6

8

10

2 4 6 8 10

c. j(x) 5 2(x 1 1)2 23; vertically stretched by a factor of 2, and translated 1 unit left and 3 units down

x

j(x)

22242628210

210

28

26

24

22

2

4

6

8

10

2 4 6 8 10


d. k(x) 5 3(x 2 2)2 2 3; vertically stretched by a factor of 3, and translated 2 units right and 3 units down

x

k(x)

22242628210

210

28

26

24

22

2

4

6

8

10

2 4 6 8 10

77. (23, 5). Sample explanation: In the vertex form of the equation, the value of h is 23 and the value of k is 2.

78. a. vertex form: h(x) 5 (x 1 2)2 2 4

vertex: (22, 24)

axis of symmetry: x 5 22

graph opens: up

b. vertex form: h(x) 5 (x 2 4)2 1 3

vertex: (4, 3)


graph opens: up

c. vertex form: h(x) 5 22(x 2 3)2 1 1

vertex: (3, 1)


graph opens: down

d. vertex form: h(x) 5 3(x 1 4)2 2 8

vertex: (24, 28)


graph opens: up

79. A

80. Sal should write 1 in the first box because adding 1 completes the square for the quadratic expression x2 2 2x inside the parentheses. Write 3 in the second box because subtracting 3 outside the parentheses keeps the expression on the right side of the function balanced.

LeSSon 12-1 81. The value of f (x) increases for x , 3 and decreases

for x . 3. Sample explanation: If the graph opens downward, the vertex is the highest point. So, the value of f (x) increases as the graph moves from left to right toward the vertex. The x-coordinate of the vertex is 3, so the function increases for x , 3. The value of f (x) decreases as the graph moves from left to right away from the vertex, so the function decreases for x . 3.

82. Sample explanation: The equation is written in standard form, ax2 1 bx 1 c 5 0, so I would use

the formula x 5 2ba2

. Substituting 212 for b and 2

for a, I find that x 5 3. Substituting 3 for x in the original equation, I find that f (x) is 9 when x is 3. The vertex is (3, 9).

83. a. (25, 210)

b. (7, 217)

c. (3, 1)

d. (21, 3)

84. B

85. a. G(x) 5 22x2 1 20x

b.

x

G(x)

210220230240250

250

240

230

220

210

10

20

30

40

50

10 20 30 40 50


c. (5, 50); the x-coordinate of 5 represents the width in feet that gives the greatest area. The y-coordinate of 50 represents the greatest area in square feet.

d. 5 ft by 10 ft; the x-coordinate of the vertex, 5, is the width that results in the greatest area. Substituting this value of x into the expression for the length gives the length that results in greatest area: 20 2 2(5) 5 10 ft.

LeSSon 12-2 86. a. x-intercepts: 2 and 5

y-intercept: 10

b. x-intercepts: 26 and 22

y-intercept: 12

c. x-intercepts: 232

and 5

y-intercept: 215

d. x-intercepts: 23 and 2

y-intercept: 224

87. An x-intercept is the x-coordinate of a point where a graph intersects the x-axis. Quadratic functions can have 0, 1, or 2 x-intercepts.

88. C

89. The graph of a quadratic function has only one x-intercept when the vertex of the graph is on the x-axis.

90 . a. 2225; the y-intercept represents the profit the tour company would make for selling the tickets for 0 each. The y-intercept is negative, which indicates a loss of money. The tour company would lose 225 if it gave the tickets away for free.

b. 5 and 45; the x-intercepts represent selling prices that would result in a profit of 0. The tour company would break even but make no profit if it were to sell the tickets for 5 or for 45.

c. Reasonable domain: 5 # x # 45; reasonable range: 0 # y # 400. Sample explanation: A graph of T(x) shows that the tour company’s profit is greater than or equal to 0 when the selling price x is between 5 and 45, so the reasonable domain is 5 # x # 45. The graph also shows that the maximum value of the tour company’s profit is 400. Because the profit must be greater than or equal to 0, the reasonable range of the function is 0 # y # 400.

d. 25; the graph of T(x) opens downward and its vertex is (25, 400). The vertex indicates that the touring company will make a maximum profit of 400 by selling the tickets for 25.

LeSSon 12-3

91. a. vertex: (74 , 298 )

y-intercept: 5

x-intercept(s): 1 and 32


minimum: 298

x

f(x)

22242628210

210

28

26

24

22

2

4

6

8

10

2 4 6 8 10


b. vertex: (1, 4)

y-intercept: 3

x-intercept(s): 21 and 3


maximum: 4

x

y

22242628210

210

28

26

24

22

2

4

6

8

10

2 4 6 8 10

92. a. The ticket price of either 15 or 35 would result in a profit of 300. Sample explanation: Set T(x) equal to 300; 300 5 2x2 1 50x 2 225. Subtract 300 from both sides to get 0 5 2x2 1 50x 2 525. 0 5 2(x 2 15)(x 2 35). So, x 5 15 or x 5 35, which means the selling price of 15 and 35 will result in a profit of 300.

b. No. The graph shows that the vertex of the profit function is (25, 400), so the maximum profit the tour company can earn is 400. To confirm, set f (x) equal to 500: 500 5 2x2 1 50x 1 225. Subtract 500 from both sides to write the equation in standard form: 0 5 2x2 1 50x 2 725. Use the Quadratic Formula to solve for x, which shows that x 5 25 6 10i. Because the equation has complex solutions, there is no real value of x that results in a profit of 500.

c. 175. Sample explanation: Evaluate T(x) for x 5 10: T(10) 5 2(10)2 1 50(10) 2 225 5 175. The tour company can expect to make a profit of 175 if it sells tickets for 10 each.

93. The y-intercept occurs when the graph intersects the y-axis at x 5 0. Evaluate the function for x 5 0. f (0) 5 (0)2 2 60(0) 1 35 5 35. The y-intercept is 35.

94. D

95. The axis of symmetry is the vertical line through the vertex, so the x-coordinate of the vertex will give you the axis of symmetry.

LeSSon 12-4 96. The value of the discriminant reveals the number

of x-intercepts of the graph of the function: when the discriminant is zero, there is one x-intercept; when the discriminant is positive, there are two x-intercepts; when the discriminant is negative, there are no x-intercepts.

97. There are two real, irrational solutions and two x-intercepts.

98. a. discriminant: 29

two real, irrational solutions

x-intercepts: 24.2 and 1.2

210 25 105

25

5

10

210

x

y


b. discriminant: 0 one real, rational solution x-intercept: 2

210 25 105

25

5

10

210

x

y

c. discriminant: 231

two complex conjugate solutions

no real solutions

no x-intercepts

210 25 105

25

5

10

210

x

y

d. discriminant: 64

two real, rational solutions

x-intercepts: 23 and 213

210 25 105

25

5

10

210

x

y

99. C

100. The discriminant is less than zero.

LeSSon 12-5 101. a.

210 25 105

5

10

x

y

25

210


b.

c.

x

y

210

25

25210 5 10

10

5

d.

102. D

103. B

104.

x

y

22242628210

28

26

24

22

2

4

6

8

10

2 4 6 8 10

210

a. not a solution

b. not a solution

c. solution

d. solution

105. a. f (DBH) 5 2r; sample explanation: The equation is linear because the diameter is two times the radius, which is a linear measure.

b. f (a) 5 10 # pr2 # 30; sample explanation: The inequality is quadratic because the area is a function of the square of the radius.

x

y

210

25

10525210

5

10

210 25 105

5

10

x

y

25

210


LeSSon 13-1 106. a.

solutions: (21.6, 1.4) and (2.6, 5.6)

b.

x

y

22242628210

210

28

26

24

22

2

4

6

8

10

2 4 6 8 10

solutions: (22, 21) and (3, 26)

c.

x

y

22242628210

210

28

26

24

22

2

4

6

8

10

2 4 6 8 10

solutions: (2.5, 4.5) and (6.5, 8.5)

d.

x

y

22242628210

210

28

26

24

22

2

4

6

8

10

2 4 6 8 10

solution: (3, 1)

x

y

22242628210

210

28

26

24

22

2

4

6

8

10

2 4 6 8 10


107. Mari is correct. If the line touches the curve at exactly one point, the system will have only one solution. An example of a system with only one

solution is y

y x(

52

5 2 2

34) 32 .

x

y

22242628210

210

28

26

24

22

2

4

6

8

10

2 4 6 8 10

108. B

109. a. The demand might increase if he lowers wthe price.

b. He will only realize a profit if the demand function is greater than the supply function.

c. The break-even point will be the point where the graphs of the demand and supply functions intersect.

110.

solutions: (27.8, 778) and (55.8, 142)

LeSSon 13-2 111. Check students’ work. Possible answers: The

linear equation is solved for y. Substitute for y in the quadratic equation. Next write the resulting equation in standard form and solve for x. Then substitute the value for x into the original equation(s) to find the corresponding value of y. Substitute the resulting values for both x and y into each of the original equations to check.

112. a. (2, 21), (25, 222)

b. (4, 17)

c. (5, 9), (2, 23)

d. no real solutions

113. B

114. The x- and y-coordinates of the two solutions are the same. The system of equations has one real solution.

115. The graphs of the equations intersect at two points, (5, 9) and (2, 23). The system has two real solutions.

x

y

22242628210

210

28

26

24

22

2

4

6

8

10

2 4 6 8 10

x

y

2202402602802100

21000

2800

2600

2400

2200

200

400

600

800

1000

20 40 60 80 100

Answers to Algebra 2 Unit 2 Practicepshs.psd202.org/documents/bmccormi/1508950153.pdfSpringBoard...

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