Inequalities Part 5 Quiz

(x − 3)² ≥ 4
Take square roots (both sides non-negative, inequality preserved):
|x − 3| ≥ 2 this means
x − 3 ≥ 2 or x − 3 ≤ −2
Case 1: x − 3 ≥ 2 → x ≥ 5
Case 2: x − 3 ≤ −2 → x ≤ 1
x ≤ 1 or x ≥ 5

|x| ≥ x − 2
Subtract x from both sides: |x| − x ≥ −2
This means:
Case 1: x − x ≥ −2 → 0 ≥ −2 (always true)
Case 2: −x − x ≥ −2 → −2x ≥ −2 → x ≤ 1
Solution: x ≤ 1

(x² − 9)/(x + 3) < 0 or, (x − 3)(x + 3)/(x + 3) < 0 For x = −3, expression behaves like Solve x − 3< 0 or, x < 3 Exclude x=−3 (denominator zero) Final: x < 3, x ≠ −3

|x + 4| ≤ |x − 2|
Square both sides (valid since both non-negative): (x + 4)² ≤ (x − 2)²
x² + 8x + 16 ≤ x² − 4x + 4
x ≤ −1

|x² − 1| ≥ 3
Case 1: x² − 1 ≥ 3 → x² ≥ 4 → x ≤ −2 or x ≥ 2
Case 2: x² − 1 ≤ −3 → x² ≤ −2 (no real solutions)
Solution: x ≤ −2 or x ≥ 2

(x − 1)/(x² − 1) > 0
or, (x − 1)/(x − 1)(x + 1) > 0
Critical points: x = -1, x = 1
Test intervals:
for x < -1 => neg <0 (false) for -1 < x < 1 => pos>0 (true)
for x > 1 => pos>0 (true)
Solution: −1 < x < 1 or x > 1

|x| > |x − 4|
Square both sides (valid since both are non-negative): x² > (x − 4)²
x² > x² − 8x + 16
8x > 16
x > 2

Numerator is always positive. Fraction is positive only when denominator is positive. So x − 5 > 0 → x > 5. Expression is undefined at 5.

Factorizing gives (2x + 3)(x − 1) ≥ 0. Since parabola opens upward, inequality holds outside roots. So solution is x ≤ −3/2 or x ≥ 1.

3 − 2x > 7 → −2x > 4. Dividing by −2 reverses inequality → x < −2.