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

A2 2015 College Board. All rights reserved. SpringBoard Algebra 2, Unit 2 Practice

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 , 1

2 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

A3 2015 College Board. All rights reserved. SpringBoard Algebra 2, Unit 2 Practice

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

A4 2015 College Board. All rights reserved. SpringBoard Algebra 2, Unit 2 Practice

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 5 136

6

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 6532

6

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

A5 2015 College Board. All rights reserved. SpringBoard Algebra 2, Unit 2 Practice

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