Ï

Chapter 4,
continued
Chapter 4 Review (pp. 318–322)
2. J;
Price (dollars/drum) +
1. To determine whether a function has a maximum value
Number sold each month (drums) > $6500
(120 2 5x) + (50 1 4x) > 6500
3. C;
2. A pure imaginary number is a complex number a 1 bi
h 5 216t 2 1 v0 t 1 h0
where a 5 0 and b Þ 0.
3 5 216t 2 1 50t 1 5
3. A function of the form y 5 a(x 2 h)2 1 k is written in
0 5 216t 2 1 50t 1 2
vertex form.
}}
250 6 Ï50 2 4(216)(2)
2(216)
4. Sample answer: y 5 4x 2 2 2x 1 7
250 6 Ï 2628
5. y 5 x 2 1 2x 2 3
2
t 5 }}
b2 2 4ac 5 (22)2 2 4(4)(7) 5 2108
}
t 5 }}
232
b
x 5 2}
5 2}
5 21
2a
2(1)
Reject the negative solution, 20.04. The ball is in the air
for about 3.16 seconds.
y 5 (21)2 1 2(21) 2 3 5 24
}
Vertex: (21, 24)
y
x 5 21
Axis of symmetry : x 5 21
}
2
2
{24 1 5i{ 5 Ï(24) 1 5 5 Ï41
2
x 5 1:
5. C;
2
y 5 1 1 2(1) 2 3 5 0; (1,0)
2
x-intercept: (0, 0), (5, 0)
x
(21, 24)
y 5 a(x 2 p)(x 2 q)
6. y 5 23x 2 1 12x 2 7
y 5 a(x 2 0)(x 2 5)
212
b
y 5 ax(x 2 5)
x 5 2}
5}
52
2a
2(23)
Use the point (3, 3.8) to find a.
y 5 23(2)2 1 12(2) 2 7 5 5
3.8 5 a(3)(3 2 5)
Vertex: (2, 5)
3.8 5 26a
y
(2, 5)
Axis of symmetry: x 5 2
20.63 ø a
Copyright © by McDougal Littell, a division of Houghton Mifflin Company.
2
t ø 3.16 or t ø 20.04
4. H;
y-intercept: 27; (0, 27)
y 5 20.63x (x 2 5)
The equation y 5 20.63x (x 2 5) models the parabolic
cross section of the lamp.
x52
x 5 1:
1
21
y 5 23(1) 1 12(1) 2 7
2
x
5 2; (1, 2)
6. J;
(5 2 9i )(5 1 9i ) 5 25 1 45i 2 45i 2 81i
2
5 25 2 81(21)
5 106
7.
or a minimum value, look at the coefficient a of the x 2
term. If a < 0, the function has a maximum value. If
a > 0, the function has a minimum value.
Area of
Area of
Area of
5 vertical 1 horizontal
stripes
stripe
stripe
1
} (8)(5) 5 8x 1 (5 2 x)(x)
3
13.33 5 8x 1 5x 2 x 2
x 2 2 13x 1 13.33 5 0
7. f (x) 5 2x 2 2 2x 2 6
b
(22)
x 5 2}
5 2}
5 21
2a
2(21)
f (21) 5 2(21)2 2 2(21) 2 6 5 25
Vertex: (21, 25)
Axis of symmetry: x 5 21
y-intercept: 26; (0, 26)
x 5 1:
}
5 29; (1, 29)
8. y 5 (x 2 1)(x 1 5)
}}
13 6 Ï(213) 2 4(1)(13.33)
2(1)
2
x 5 }}}
}
13 6 Ï 115.68
x 5 }}
2
y
2
22
x
(21, 25)
f (1) 5 212 2 2(1) 2 6
x 5 }}
2a
2b 6 Ïb 2 2 4ac
x 5 21
x 5 22
x-intercepts: p 5 1 and q 5 25
p1q
y
2
21
1 1 (25)
x
x5}
5}
5 22
2
2
y 5 (22, 21)(22 1 5) 5 29
x ø 11.9 or x ø 1.1
Vertex: (22, 29)
If x ø 11.9 inches, the border will be bigger than the
paper, so x ø 1.1 inches. The width x of the stripes will
be about 1.1 inches.
Axis of symmetry: x 5 22
(22, 29)
Algebra 2
Worked-Out Solution Key
249
Chapter 4,
continued
9. g(x) 5 (x 1 3)(x 2 2)
1
1
21
x
Because the x-intercepts are 0 and 33, the flea jumped a
distance of 33 centimeters.
p1q
1
1
2
x 5 22
1
4
(2 , 26 )
y 5 20.073(16.5)(16.5 2 33) ø 19.9
25
5 2}
4
The flea’s maximum height was about 19.9 centimeters.
Vertex: 1 2}2 , 2}
42
25
1
15. x 2 1 5x 5 0
x(x 1 5) 5 0
1
Axis of symmetry: x 5 2}2
x 5 0 or x 1 5 5 0
10. y 5 23(x 1 1)(x 2 6)
y
1
22,
(
x-intercepts: p 5 21 and q 5 6
p1q
21 1 6
3
36 4
x 5 0 or
)
5
z 2 2 63z 5 0
z(z 2 63) 5 0
y 5 231 }2 1 1 21 }2 2 6 2 5 }
4
5
147
5
5 147
Vertex: }2 , }
4
x 5 25
z2 5 63z
16.
x5}
5}
5 }2
2
2
1
0 1 33
5}
5 16.5
x5}
2
2
2
z50
1
x 5 22
5
or z 2 63 5 0
z 5 0 or
22
x
5
17.
Axis of symmetry: x 5 }2
z 5 63
s 2 6s 2 27 5 0
2
(s 1 3)(s 2 9) 5 0
11. y 5 (x 2 2)2 1 3
s1350
y
or s 2 9 5 0
s 5 23 or
Vertex: (2, 3)
18. k 1 12k 2 45 5 0
Axis of symmetry: x 5 2
(k 1 15)(k 2 3) 5 0
x 5 0:
y 5 (0 2 2) 1 3 5 7; (0, 7)
2
x 5 1:
y 5 (1 2 2) 1 3 5 4; (1, 4)
k 1 15 5 0
(2, 3)
1
2
x52
21
x
12. f (x) 5 (x 1 6) 1 8
x 1 18x 5 281
19.
x 2 1 18x 1 81 5 0
y
(x 1 9)2 5 0
x1950
Axis of symmetry: x 5 26
x 5 22:
x 5 29
(26, 8)
f (22) 5 (22 1 6) 1 8
2
x 5 24:
n2 1 5n 5 24
20.
x 5 26
5 24; (22, 24)
n2 1 5n 2 24 5 0
2
21
(n 1 8)(n 2 3) 5 0
x
n1850
f (24) 5 (24 1 6) 1 8 5 12; (24, 12)
2
x 5 28
Vertex: (28, 23)
21.
Original
playground
y 5 22(26 1 8) 2 3
2
x 5 25:
y 5 22(25 1 8)2 2 3 5 221; (25, 221)
n53
x
x
(28, 23)
5 211; (26, 211)
or n 2 3 5 0
n 5 28 or
y
1
21
Axis of symmetry: x 5 28
x 5 26:
k53
2
Vertex: (26, 8)
13. y 5 22(x 1 8) 2 33
or k 2 3 5 0
k 5 215 or
2
2
s59
2
x
48 ft
72 ft
New
New
New area
5 length + width
(square feet)
(feet)
(feet)
2(72)(48)
5 (72 1 x) + (48 1 x)
6912 5 3456 1 120x 1 x 2
0 5 x 2 1 120x 2 3456
0 5 (x 1 144)(x 2 24)
250
Algebra 2
Worked-Out Solution Key
Copyright © by McDougal Littell, a division of Houghton Mifflin Company.
1
y 5 20.073(x 2 0)(x 2 33)
2
p1q
23 1 2
1
x5}
5}
5 2}2
2
2
g1 2}2 2 5 1 2}2 1 3 21 2}2 2 2 2
14. y 5 20.073x(x 2 33)
y
x-intercepts: p 5 23 and q 5 2
Chapter 4,
continued
x 1 144 5 0
or x 2 24 5 0
x 5 2144 or
35. x 2 2 6x 2 15 5 0
x2 2 6x 5 15
x 5 24
x 2 2 6x 1 9 5 15 1 9
Reject the negative value, 2144. The playground’s length
and width should each be increased by 24 feet.
16 5 38r 2 12r
22.
(x 2 3)2 5 24
2
12r 2 2 38r 1 16 5 0
6r 2 19r 1 8 5 0
36. 3x 2 2 12x 1 1 5 0
(3r 2 8)(2r 2 1) 5 0
3r 2 8 5 0
8
r 5 }3
or
2r 2 1 5 0
or
1
r 5 }2
3x2 2 12x 5 21
3(x2 2 4x) 5 21
3(x 2 4x 1 4) 5 21 1 3(4)
2
3(x 2 2)2 5 11
23. 3x 2 2 24x 2 48 5 0
x 2 2 8x 2 16 5 0
11
(x 2 2)2 5 }
3
}}
2(28) 6 Ï(28) 2 4(1)(216)
2
x 5 }}}
2(1)
Î
}
11
x2256 }
3
}
8 6 Ï 128
5}
2
Î
}
11
x526 }
3
}
8 6 8Ï 2
5}
2
}
Ï33
x526}
3
}
5 4 6 4Ï 2
}
x 5 4 2 Ï2
or
}
x 5 4 1 Ï2
37. x 2 1 3x 2 1 5 0
24. 20a 2 13a 2 21 5 0
x2 1 3x 5 1
2
(4a 1 3)(5a 2 7) 5 0
9
4a 1 3 5 0
3
1 x 1 }32 2
2
7
a 5 }5
25. 3x 2 5 108
26. 5y 2 1 4 5 14
}
y2 5 2
27. 3(p 1 1)2 5 81
3
}
s 5 4:r 2
28.
(p 1 1) 5 27
510,000,000 5 4:r
3
p 1 1 5 6Ï27
}}
}
24 6 Ï 28
Ï127,500,000
6 }}
5r
:F
}
p 5 21 6 3Ï3
66370.6 5 r
The radius of Earth is
about 6370.6 kilometers.
29. 29i(2 2 i) 5 218i 1 9i 2 5 218i 1 9(21) 5 29 2 18i
30. (5 1 i)(4 2 2i) 5 20 2 10i 1 4i 2 2i 2
31. (2 2 5i)(2 1 5i) 5 4 1 10i 2 10i 2 25i2 5 4 2 25(21)
5 4 1 25 5 29
}
}
The solutions}are x 5 22 6 Ï 7 ø 0.65 and
x 5 22 2 Ï 7 ø 24.65.0.
9x 2 5 26x 2 1
39.
9x 1 6x 1 1 5 0
2
}}
26 6 Ï62 2 4(9)(1)
x 5 }}
2(9)
}
5 20 2 6i 2 2(21) 5 22 2 6i
}
24 6 2Ï 7
x5}
5}
5 22 6 Ï 7
2
2
}}
}
}
Ï13
24 6 Ï42 2 4(1)(23)
x 5 }}
2(1)
2
}
5 r2
:
p 1 1 5 63Ï3
13
38. x 2 1 4x 2 3 5 0
127,500,000
}
Ï
x 5 2}2 6 }
2
y 5 6Ï 2
x 5 66
}
3
2
x 5 6 Ï 36
13
5}
4
x 1 }2 5 6 }
4
5y 5 10
2
9
x2 1 3x 1 }4 5 1 1 }4
or 5a 2 7 5 0
a 5 2}4 or
Copyright © by McDougal Littell, a division of Houghton Mifflin Company.
}
x 5 3 6 2Ï 6
2
x 5 36
}
x 2 3 5 6Ï 24
26 6 Ï 0
26
1
5}
5 2}3
x5}
18
18
1
The solution is 2}3 .
32. (8 2 6i) 1 (7 1 4i) 5 (8 1 7) 1 (26 1 4)i 5 15 2 2i
33. (2 2 3i) 2 (6 2 5i) 5 (2 2 6) 1 (23 1 5)i 5 24 1 2i
4i
212i 2 24i2
23 2 6i
4i
34. } 5 } + } 5 }}2
23 1 6i 23 2 6i
23 1 6i
9 1 18i 2 18i 2 36i
212i 2 24(21)
24 2 12i
8
4
5 }}
5}
5}
2}
i
45
15
15
9 2 36(21)
Algebra 2
Worked-Out Solution Key
251
Chapter 4,
continued
1
44. } x 2 1 3x 2 6 > 0
2
6x2 2 8x 5 23
6x 2 8x 1 3 5 0
2
1
2
2(28) 6 Ï(28) 2 4(6)(3)
2
x 5 }}}
2(6)
8 6 Ï28
}
}
Î
}}
23 6 32 2 41 }2 2(26)
1
x5
4 6 iÏ2
x5}
5}
5}
12
12
6
4 1 iÏ 2
21 }2 2
1
}
}
4 2 iÏ 2
x 5 23 6 Ï21
The solutions are }
and }
.
6
6
x ø 1.58 or x ø 27.58
41. h 5 216t 2 1 v0t 1 h0
The solution of the inequality is approximately
x < 27.58 or x > 1.58.
0 5 216t 2 2 40t 1 9
}}
}
2(240) 6 Ï(240)2 2 4(216)(9)
2(216)
45.
40 6 Ï2176
232
t 5 }}} 5 }
y 5 a(x 2 p)(x 2 q)
y 5 a(x 1 3)(x 2 2)
t ø 22.71 or t ø 0.21
12 5 a(3 1 3)(3 2 2)
Reject the negative solution. The ball is in the air for about
0.21 second after it is spiked.
12 5 6a
42. 2x 2 2 11x 1 5 < 0
25a
A quadratic function is y 5 2(x 1 3)(x 2 2).
y
2
2x2 2 11x 1 5 5 0
2
(2x 2 1)(x 2 5) 5 0
x
46. y 5 ax 2 1 bx 1 c
2 5 a(5)2 1 b(5) 1 c l 25a 1 5b 1 c 5 2
1
x 5 }2 or x 5 5
2 5 a(0)2 1 b(0) 1 c l
c52
26 5 a(8)2 1 b(8) 1 c l 64a 1 8b 1 c 5 26
The solution of the inequality
25a 1 5b 1 c 5 2
1
is }2 < x < 5.
25a 1 5b 1 2 5 2
25a 1 5b 5 0
43. 2x 1 4x 1 3q0
2
64a 1 8b 1 c 5 26
2x 1 4x 1 3 5 0
2
64a 1 8b 1 2 5 26
}}
24 6 Ï42 2 4(21)(3)
2(21)
x 5 }}
}
24 6 Ï28
64a 1 8b 5 28
38
200a 1 40b 5 0
3 (25)
2320a 2 40b 5 40
25a 1 5b 5 0
}
24 6 2Ï 7
x5}
5}
22
22
64a 1 8b 5 28
}
5 2 6 Ï7
2120a 5 40
x ø 4.65 or x ø 20.65
1
a 5 2}3
y
25a 1 5b 5 0
25 1 2}3 2 1 5b 5 0
1
25
3
1 5b 5 0
2}
3
25
1
5b 5 }
3
x
5
b 5 }3
The solution of the inequality is approximately
5
1
The solution is a 5 2}3, b 5 }3 , and c 5 2. A quadratic
20.65axa4.65.
1
5
function is y 5 2}3 x2 1 }3 x 1 2.
47.
y 5 a(x 2 h)2 1 k
y 5 a(x 2 2)2 1 7
2 5 a(4 2 2)2 1 7
2 5 4a 1 7
25 5 4a
5
2}4 5 a
5
A quadratic function is y 5 2}4 (x 2 2)2 1 7.
252
x
}}
}
8 6 2Ï 2 i
22
} x 2 1 3x 2 6 5 0
}}
}
y
2
Algebra 2
Worked-Out Solution Key
Copyright © by McDougal Littell, a division of Houghton Mifflin Company.
40.
Chapter 4,
continued
y 5 a(x 2 h)2 1 k
48.
x 5 21
y 5 a(x 2 12) 1 7
2
y
2
22
0 5 a(0 2 12)2 1 7
x
0 5 144a 1 7
27 5 144a
7
2}
5a
144
A quadratic function that models the soccer ball’s path is
(21, 218)
7
y 5 2}
(x 2 12)2 1 7.
144
4. x 2 11x 1 30 5 (x 2 6)(x 2 5)
2
5. z 2 1 2z 2 15 5 (z 1 5)(z 2 3)
Chapter 4 Test (p. 323)
6. n 2 2 64 5 n2 2 82 5 (n 1 8)(n 2 8)
1. y 5 x2 2 8x 2 20
7. 2s 2 1 7s 2 15 5 (2s 2 3)(s 1 5)
(28)
b
x 5 2}
5 2}
54
2a
2(1)
8. 9x 2 1 30x 1 25 5 3x 2 1 2(3x)(5) 1 52 5 (3x 1 5)2
9. 6t 2 1 23t 1 20 5 (3t 1 4)(2t 1 5)
y 5 42 2 8(4) 2 20 5 236
10.
Vertex: (4, 236)
(x 1 5)(x 2 8) 5 0
Axis of symmetry: x 5 4
x1550
y-intercept: 220; (0, 220)
or x 2 8 5 0
x 5 25 or
x 5 22:
y 5 (22) 2 8(22) 2 20 5 0; (22, 0)
2
(r 2 7)(r 2 6) 5 0
y
x
x54
x58
11. r 2 2 13r 1 42 5 0
5
21
x2 2 3x 2 40 5 0
12.
r2750
or
r2650
r57
or
r56
2w 2 1 13w 2 7 5 0
(2w 2 1)(w 1 7) 5 0
Copyright © by McDougal Littell, a division of Houghton Mifflin Company.
2w 2 1 5 0 or w 1 7 5 0
(4, 236)
2. y 5 2(x 1 3) 1 5
1
w 5 }2 or
2
13. 10y 2 1 11y 2 6 5 0
Vertex: (23, 5)
Axis of symmetry: x 5 23
x 5 22: y 5 2(22 1 3)2 1 5 5 4; (22, 4)
(5y 2 2)(2y 1 3) 5 0
5y 2 2 5 0
or
2
y 5 }5
or
x 5 21: y 5 2(21 1 3)2 1 5 5 1; (21, 1)
(23, 5) y
14. 2(m 2 7)2 5 16
(m 2 7)2 5 8
x 5 23
x
3. f(x) 5 2(x 1 4)(x 2 2)
x-intercepts: p 5 24 and q 5 2
p1q
24 1 2
5}
5 21
x5}
2
2
f(21) 5 2(21 1 4)(21 2 2) 5 218
Vertex: (21, 218)
2y 1 3 5 0
3
y 5 2}2
15. (x 1 2)2 2 12 5 36
(x 1 2)2 5 48
}
m 2 7 5 6 Ï8
1
21
w 5 27
}
x 1 2 5 6Ï48
}
}
x 5 22 6 4Ï3
m 5 7 6 2Ï 2
16. (3 1 4i) 2 (2 2 5i) 5 (3 2 2) 1 (4 1 5)i 5 1 1 9i
17. (2 2 7i)(1 1 2i) 5 2 1 4i 2 7i 2 14i 2
5 2 2 3i 2 14(21) 5 16 2 3i
3 1 i 2 1 3i
6 1 9i 1 2i 1 3i2
31i
18. } 5 } + } 5 }}2
2 2 3i 2 1 3i
2 2 3i
4 1 6i 2 6i 2 9i
6 1 11i 1 3(21)
3 1 11i
3
11
5}
5}
1}
i
5 }}
13
13
13
4 2 9(21)
Axis of symmetry: x 5 21
Algebra 2
Worked-Out Solution Key
253
Chapter 4,
continued
19. x 2 1 4x 2 14 5 0
20. x 2 2 10x 2 7 5 0
25. yqx 2 2 8
x 1 4x 5 14
x 2 10x 5 7
Test (0, 0).
x2 1 4x 1 4 5 14 1 4
x2 2 10x 1 25 5 7 1 25
yqx2 2 8
2 0q
2
(x 1 2)2 5 18
2
(x 2 5)2 5 32
}
}
x 1 2 5 6Ï18
x 2 5 5 6Ï 32
x 1 2 5 63Ï2
x 2 5 5 64Ï 2
}
}
}
x 5 22 6 3Ï2
}
22 1 3Ï2 and
and 5 2 4Ï 2 .
}
x
q2
26. y < x2 1 4x 2 21
y
3
Test (0, 0).
22 2 3Ï2 .
21.
21
}
The solutions are 5 1 4Ï 2
}
(0, 0)
}
x 5 5 6 4Ï 2
The solutions are
y
1
21
y < x2 1 4x 2 21
0
< 02 1 4(0) 2 21
4x 1 8x 1 3 5 0
2
4x2 1 8x 5 23
(0, 0)
x
0 ñ 221 4(x2 1 2x) 5 23
4(x2 1 2x 1 1) 5 23 1 (4)(1)
4(x 1 1)2 5 1
27. y > 2x 2 1 5x 1 50
1
(x 1 1)2 5 }4
y
Test (0, 0).
}
Ï
1
x 1 1 5 6 }4
y > 2x2 1 5x 1 50
0
> 202 1 5(0) 1 50
1
x 1 1 5 6 }2
0 ò 50 1
x 5 21 6 }2
7
3
1
The solutions are 2}2 and 2}2.
28.
y 5 a(x 2 p)(x 2 q)
12 5 12a
}
210 6 4Ï 10
15a
}
25 6 2Ï10
x5}
5}
5}
6
3
6
A quadratic function is y 5 (x 1 7)(x 1 3).
}
25 1 2Ï 10
29.
The solutions are x 5 }
ø 0.44 and
3
y 5 a(x 1 3)2 2 2
}
25 2 2Ï10
210 5 a(1 1 3)2 2 2
ø 23.77.
x5}
3
210 5 16a 2 2
23. 2x 2 2 x 1 6 5 0
}}
2(21) 6 Ï(21)2 2 4(2)(6)
1 6 iÏ47
1 6 Ï 247
x 5 }}}
5}
5}
4
2(2)
4
}
}
}
28 5 16a
1
2}2 5 a
}
1 1 iÏ 47
1 2 iÏ 47
1
and }
.
The solutions are }
4
4
A quadratic function is y 5 2}2 (x 1 3)2 2 2.
24. 5x 2 1 2x 1 5 5 0
30. y 5 ax 2 1 bx 1 c
}}
22 6 Ï22 2 4(5)(5)
y 5 a(x 2 h)2 1 k
}
22 6 Ï 296
}
22 6 4iÏ 6
5}
5}
x 5 }}
10
10
2(5)
}
21 6 2iÏ 6
8 5 a(4)2 1 b(4) 1 c l 16a 1 4b 1 c 5
0 5 a(8)2 1 b(8) 1 c l 64a 1 8b 1 c 5
x5}
5
}
21 6 2iÏ6
}
21 2 2iÏ6
The solutions are }
and }
.
5
5
16a 1 4b 1 c 5 88 3 (22)
64a 1 8b 1 c 5
64a 1 8b 1 c 5 0
16a 1 4b 1 c 5 8
49a 1 7b 1 c 5 24
3 (27)
34
0
2 c 5 216
2112a 2 28b 2 7c 5 256
196a 1 28b 1 4c 5 216
84a
Algebra 2
Worked-Out Solution Key
0
232a 2 8b 2 2c 5 216
32a
254
8
24 5 a(7)2 1 b(7) 1 c l 49a 1 7b 1 c 5 24
2 3c 5 272
Copyright © by McDougal Littell, a division of Houghton Mifflin Company.
12 5 a(21 1 7)(21 1 3)
}}
210 6 Ï10 2 2 4(3)(25)
x 5 }}
2(3)
210 6 Ï160
x
y 5 a(x 1 7)(x 1 3)
22. 3x 1 10x 2 5 5 0
2
}
(0, 0)
22
Chapter 4,
32a 2 c 5 216
continued
3 (23)
296a 1 3c 5
48
84a 2 3c 5 272
84a 2 3c 5 272
212a
5 224
a52
3. B;
V 5 :r 2h 5 :(2)2(12) 5 48:
The vase has a volume of 48: cubic inches.
4. H;
} }
RT i VX, so ŽSTU > ŽWVU by the Alternate Interior
Angles Theorem, so nSTU is similar to nWVU.
32a 2 c 5 216
32(2) 2 c 5 216
64 2 c 5 216
US
TU
UW
VU
US
16
16
20
}5}
2c 5 280
}5}
c 5 80
16a 1 4b 1 c 5 8
20US 5 256
16(2) 1 4b 1 80 5 8
US 5 12.8
}
The length of US is 12.8 units.
32 1 4b 1 80 5 8
112 1 4b 5 8
5. C;
4b 5 2104
b 5 226
72 5 h 2 1 3.52
7
h
36.75 5 h
The solution is a 5 2, b 5 226, and c 5 80. A quadratic
function is y 5 2x2 2 26x 1 80.
31. (16x)2 1 (9x)2 5 322
}
Ï36.75 5 h
3.5
3.5
6.06 ø h
Total height 5 height of roof 1 height of box
256x2 1 81x2 5 1024
ø 6.06 1 8
337x 2 5 1024
5 14.06
1024
x2 5 }
337
The birdhouse is about 14.1 inches tall.
Î1024
}
x56 }
337
6. G;
y 5 x 2 2 x 2 30
y 5 (x 2 6)(x 1 5)
x ø 6 1.74
Copyright © by McDougal Littell, a division of Houghton Mifflin Company.
7
2
Reject the negative solution. The widescreen TV has a
width of (16)(1.74) 5 27.84 inches and a height
of (9)(1.74) 5 15.66 inches.
32. The best-fitting quadratic model is
s 5 0.0008m2 2 0.048m 1 1.12.
x2650
or
x56
or
x1550
x 5 25
The x-intercepts of the graph are x 5 25 and x 5 6.
7. C;
3z 2 2 1 4z 5 2z 1 13
5z 5 15
TAKS Practice (pp. 326–327)
z53
1. B;
430 5 (1.5x) 1 x
2
184,900 5 3.25x
2
2
2
56,892.3 ø x 2
}
Ï56,892.3 ø x
238.5 ø x
1.5(238.5) 1 238.5 5 596.25
Difference 5 596.25 2 430 ø 166 feet
Elizabeth saves about 166 feet by walking diagonally
across the field.
The solution of the equation is z 5 3.
8. G;
30,000(0.035) 5 1050
The whale consumes 1050 kilograms of food per day.
9. D;
The line is solid, so choices A and C can be eliminated.
The half-plane above the line is shaded, so choice B can
1
be eliminated. The inequality y q 2}2 x 2 2 is graphed.
10. J;
5x 2 3y 5 15
2. G;
*2 5 52 1 3.52
5x 2 3(0) 5 15
x53
* 5 37.25
2
}
* 5 Ï37.25
* ø 6.1
Nate needs about 6.1 feet of rope.
The coordinates of the x-intercept are (3, 0).
11. C;
A triangular pyramid has 4 vertices, and a rectangular
prism has twice as many, or 8 vertices.
Algebra 2
Worked-Out Solution Key
255
Chapter 4,
continued
12. H;
1
A 5 }2 (b1 1 b2)h
1
5 }2 (3x 2 5 1 x 1 3)(x)
1
5 }2 (4x 2 2)(x)
5 2x 2 2 x
The expression 2x 2 2 x best represents the area of
the trapezoid.
13. D;
6x 2 10y 5 18
23x 1 5y 5 212
6x 2 10y 5 18
32
26x 1 10y 5 224
0 5 26
Because the statement 0 5 26 is never true, there is
no solution.
14. H;
ŽZPW and ŽWPY are supplementary angles, so
mŽZPW 1 mŽWPY 5 1808.
15. h 5 216t 2 2 40t 1 10
0 5 22(8t 2 1 20t 2 5)
0 5 8t 2 1 20t 2 5
}}
220 6 Ï 202 2 4(8)(25)
t 5 }}
2(8)
}
220 6 Ï 560
t5}
16
Time must be positive. So, the ball hit the ground in
about 0.23 second.
256
Algebra 2
Worked-Out Solution Key
Copyright © by McDougal Littell, a division of Houghton Mifflin Company.
t ø 0.23 or t ø 22.73