Algebra Rules Review Sheet Page 18
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18
REVIEW OF ALGEBRA
56. x
2
2
2
2
2
2
2
− 16x + 80 = [x
− 16x] + 80 = [x
− 16x + (8)
− (8)
] + 80 = (x − 8)
+ 80 − 64 = (x − 8)
+ 16
2
2
2
2
57. x
2
2
2
5
5
5
25
5
15
− 5x + 10 = [x
− 5x] + 10 = x
− 5x + −
− −
+ 10 = x −
+ 10 −
= x −
+
2
2
2
4
2
4
2
2
2
2
2
58. x
2
2
2
3
3
3
3
3
5
+ 3x + 1 = [x
+ 3x] + 1 = x
+ 3x +
−
+ 1 = x +
+ 1 −
= x +
−
2
2
2
2
2
4
2
2
2
2
59. 4x
2
2
2
1
1
1
1
1
+ 4x − 2 = 4[x
+ x] − 2 = 4 x
+ x +
−
− 2 = 4 x +
− 2 − 4
= 4 x +
− 3
2
2
2
4
2
60. 3x
2
2
2
2
2
2
2
− 24x + 50 = 3[x
− 8x] + 50 = 3[x
− 8x + (−4)
− (−4)
] + 50 = 3(x − 4)
+ 50 − 3(−4)
2
= 3(x − 4)
+ 2
61. x
2
− 9x − 10 = 0 ⇔ (x + 10)(x − 1) = 0 ⇔ x + 10 = 0 or x − 1 = 0 ⇔ x = −10 or x = 1.
62. x
2
− 2x − 8 = 0 ⇔ (x − 4)(x + 2) = 0 ⇔ x − 4 = 0 or x + 2 = 0 ⇔ x = 4 or x = −2.
√
2
−9 ±
9
− 4(1)(−1)
9 ±
85
63. Using the quadratic formula, x
2
.
+ 9x − 1 = 0 ⇔ x =
=
2(1)
2
√
√
2 ±
4 − 4(1)(−7)
2 ±
32
64. Using the quadratic formula, x
2
− 2x − 7 = 0 ⇔ x =
=
= 1 ± 2
2.
2
2
√
2
−5 ±
5
− 4(3)(1)
−5 ±
13
65. Using the quadratic formula, 3x
2
.
+ 5x + 1 = 0 ⇔ x =
=
2(3)
6
√
−7 ±
49 − 4(2)(2)
−7 ±
33
66. Using the quadratic formula, 2x
2
.
+ 7x + 2 = 0 ⇔ x =
=
2(2)
4
67. Let p(x) = x
− 2x + 1, and notice that p(1) = 0, so by the Factor Theorem, (x − 1) is a factor.
3
Use long division:
2
x
+ x
− 1
3
2
x − 1 x
+ 0x
− 2x + 1
3
2
x
− x
2
x
− 2x
2
x
− x
− x + 1
− x + 1
Therefore x
3
2
+ x − 1) = 0 ⇔ x − 1 = 0 or x
2
− 2x + 1 = (x − 1)(x
+ x − 1 = 0 ⇔
√
−1 ±
1
2
− 4(1)(−1)
−1 ±
5
x = 1 or [using the quadratic formula] x =
.
=
2(1)
2
68. Let p(x) = x
3
2
+ x − 1, and notice that p(−1) = 0, so by the Factor Theorem, (x + 1) is a factor.
+ 3x
Use long division:
2
x
+ 2x − 1
3
2
x + 1 x
+ 3x
+ x − 1
3
2
x
+ x
2
2x
+ x
2
2x
+ 2x
− x − 1
− x − 1
Therefore x
3
2
2
+ 2x − 1) = 0 ⇔ x + 1 = 0 or x
2
+ 3x
+ x − 1 = (x + 1)(x
+ 2x − 1 = 0 ⇔
√
−2 ±
2
2
− 4(1)(−1)
x = −1 or [using the quadratic formula] x =
= −1 ±
2.
2
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