☰ Contents · Mathematics

Irrational equations

Lessons 11 · 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
11

Simple irrational equations and systems

Textbook, Part 1: pp. 64–68
GoalSolve irrational equations by squaring, checking roots and substitution; know the difference between even and odd roots.
New words
irrational equation · irratsional tenglamasquaring · kvadratga ko‘tarishextraneous root · chet ildizdomain · aniqlanish sohasi
Explanation

An equation with the unknown under a root sign is an irrational equation. The equation √f(x) = g(x) is equivalent to the system {f(x) = g²(x), g(x) ≥ 0}: after squaring, the condition g(x) ≥ 0 must be checked, otherwise extraneous roots remain (an even root is never negative). The equation √f(x) = √g(x) is equivalent to f(x) = g(x) with f(x) ≥ 0. An odd root has no restriction: ³√f(x) = g(x) ⇔ f(x) = g³(x) and ³√f(x) = ³√g(x) ⇔ f(x) = g(x). The equation f(x) · √g(x) = 0 means the system f(x) = 0, g(x) ≥ 0, or the equation g(x) = 0. In harder equations the root is replaced by a letter z; in systems the domain of every root expression is taken into account (for example, if √x + √y = 5, then x ≥ 0 and y ≥ 0).

Worked examples
√(2x + 3) = x. Equivalent system: 2x + 3 = x², x ≥ 0. x² − 2x − 3 = 0 ⇒ x = 3 or x = −1; the condition x ≥ 0 removes x = −1. Check: √9 = 3. Answer: x = 3.
(x − 4)√(x + 5) = 0. First case: x − 4 = 0 and x + 5 ≥ 0 ⇒ x = 4. Second case: √(x + 5) = 0 ⇒ x = −5. Answer: x = 4 and x = −5.
Class activity

“The squaring trap”: the teacher solves √(x + 6) = x by squaring and deliberately skips the check; students find which root is wrong and prove it by substituting into the original equation.

Practice
1
Which system is √f(x) = g(x) equivalent to?
2
Solve √(x + 6) = x.
3
Solve √(3x − 2) = −4.
4
Why is x = −2, found after squaring √(x + 6) = x, not a solution?