Lessons 15 · 1 lessons · M. A. Mirzaahmedov, Sh. N. Ismailov, A. Q. Amanov (algebra and analysis), B. Q. Haydarov (geometry). Mathematics Grade 10, Parts 1 and 2, 1st edition. EXTREMUM PRESS, Tashkent, 2017
15
Simple irrational inequalities
Textbook, Part 1: pp. 79–84
GoalSolve the basic irrational inequalities √A < B, √A > B, √A > √B by reducing them to equivalent systems.
New words
irrational inequality · irratsional tengsizlikequivalent system · tengkuchli sistemaexistence condition of the root · ildizning mavjudlik shartinonnegative · nomanfiy
Explanation
An inequality with the unknown under a root sign is an irrational inequality. Since the solution set is infinite, an answer cannot be checked by substitution, so every transformation must be equivalent. Raising both sides to an even power gives an equivalent inequality only when both sides are nonnegative. Rules: √A < B ⇔ {A < B², A ≥ 0, B ≥ 0}; √A > B ⇔ either {B < 0, A ≥ 0} or {B ≥ 0, A > B²} (the union of two cases); √A > √B ⇔ {B ≥ 0, A > B}. Also √A ≤ B ⇔ {A ≥ 0, B ≥ 0, A ≤ B²}. The first inequality comes from squaring, the second from the existence of the root, the third from the right to square (B ≥ 0). If B < 0, then √A < B has no solution, while √A > B holds for every x with A ≥ 0.
Worked examples
√(x + 6) < x. System: x + 6 ≥ 0, x ≥ 0, x + 6 < x². x² − x − 6 > 0 ⇒ x < −2 or x > 3. Intersect with x ≥ 0: x > 3. Check: at x = 4, √10 ≈ 3.16 < 4. Answer: (3, ∞).
√(2x + 1) > x − 1. Case 1: x − 1 < 0 and 2x + 1 ≥ 0 ⇒ −1/2 ≤ x < 1. Case 2: x ≥ 1 and 2x + 1 > (x − 1)² ⇒ x² − 4x < 0 ⇒ 0 < x < 4, so 1 ≤ x < 4. Union: [−1/2, 4). Check: at x = 4, 3 > 3 is false; at x = 3, √7 ≈ 2.65 > 2 is true.
Class activity
“Find both cases” contest: groups get inequalities √A > B; they first split into two cases by the sign of B, solve each, and unite the results. The fastest correct group wins.
Practice
1
To which equivalent system does √A < B reduce?
{A ≥ 0, B ≥ 0, A < B²}
2
Solve √(x − 2) > −1.
[2, ∞): the right side is negative and the left side nonnegative; only x − 2 ≥ 0 remains.
3
Solve √(2x − 1) < 3.
0 ≤ 2x − 1 < 9 ⇒ [0.5, 5).
4
Why can’t √x < 2 be solved by just writing “x < 4”?
Squaring loses the existence condition x ≥ 0: x = −1 satisfies x < 4 but √(−1) does not exist. The correct answer is [0, 4).