# Algebra Rules Review Sheet Page 22

22
REVIEW OF ALGEBRA
136. x
2
2
5 ≥ 0. Case 1: x ≥
5 and x ≥ −
5, so x ∈
5, ∞ .
≥ 5 ⇔ x
− 5 ≥ 0 ⇔
x −
5
x +
Case 2: x ≤
5 and x ≤ −
5, so x ∈ −∞, −
5 . Thus, the solution set is −∞, −
5, ∞ .
5 ∪
Another method: x
2
5 or x ≤ −
≥ 5 ⇔ |x| ≥
5 ⇔ x ≥
5.
137. x
3
2
2
(x − 1) ≤ 0. Since x
2
≥ 0 for all x, the inequality is satisﬁed when x − 1 ≤ 0 ⇔ x ≤ 1.
− x
≤ 0 ⇔ x
Thus, the solution set is (−∞, 1].
138. (x + 1)(x − 2)(x + 3) = 0 ⇔ x = −1, 2, or −3. Construct a chart:
Interval
x + 1
x − 2
x + 3
(x + 1)(x − 2)(x + 3)
x < −3
−3 < x < −1
+
+
−1 < x < 2
+
+
x > 2
+
+
+
+
Thus, (x + 1)(x − 2)(x + 3) ≥ 0 on [−3, −1] and [2, ∞), and the solution set is [−3, −1] ∪ [2, ∞).
139. x
3
3
2
− 1 > 0 ⇔ x(x − 1)(x + 1) > 0. Construct a chart:
> x ⇔ x
− x > 0 ⇔ x x
Interval
x
x − 1
x + 1
x(x − 1)(x + 1)
x < −1
−1 < x < 0
+
+
0 < x < 1
+
+
x > 1
+
+
+
+
Since x
3
> x when the last column is positive, the solution set is (−1, 0) ∪ (1, ∞).
140. x
3
2
3
2
2
+ 3x < 4x
⇔ x
− 4x
+ 3x < 0 ⇔ x x
− 4x + 3 < 0 ⇔ x(x − 1)(x − 3) < 0.
Interval
x
x − 1
x − 3
x(x − 1)(x − 3)
x < 0
0 < x < 1
+
+
1 < x < 3
+
+
x > 3
+
+
+
+
Thus, the solution set is (−∞, 0) ∪ (1, 3).
141. 1/x < 4. This is clearly true for x < 0. So suppose x > 0. then 1/x < 4 ⇔ 1 < 4x ⇔
1
< x. Thus, the
4
solution set is (−∞, 0) ∪
1
, ∞ .
4