Irrational inequalities
Answers are for parents and teachers.
1
Solve √(4x + 1) ≤ 5.
0 ≤ 4x + 1 ≤ 25 ⇒ [−0.25, 6].
2
Solve √(x − 3) > 2.
x − 3 > 4 ⇒ x > 7 ⇒ (7, ∞).
3
Solve √(x + 2) > x.
Case 1: x < 0, x + 2 ≥ 0 ⇒ [−2, 0). Case 2: x ≥ 0, x + 2 > x² ⇒ [0, 2). Union: [−2, 2).
4
Solve √(x + 4) ≤ x + 2.
x ≥ −2 and x + 4 ≤ x² + 4x + 4 ⇒ x² + 3x ≥ 0 ⇒ x ≥ 0 ⇒ [0, ∞).
5
Why is the solution of √(x − 5) > −2 simply x ≥ 5?
A root is always nonnegative, hence greater than −2; so the inequality holds wherever the root exists, i.e. for x − 5 ≥ 0.