Irrational equations
Answers are for parents and teachers.
1
Solve √(5x − 1) = 3.
5x − 1 = 9 ⇒ x = 2; 3 ≥ 0 holds. Answer: 2.
2
Solve √(x² − 3x + 1) = x − 1.
x² − 3x + 1 = x² − 2x + 1 ⇒ x = 0; but x − 1 = −1 < 0. No solution.
3
Solve (x² − x − 6)√(x − 1) = 0.
x² − x − 6 = 0 ⇒ 3, −2, with x ≥ 1 ⇒ 3; √(x − 1) = 0 ⇒ 1. Answer: 3 and 1.
4
Solve ³√(x² − 4) = ³√(3x).
x² − 4 = 3x ⇒ x² − 3x − 4 = 0 ⇒ x = 4 or −1; an odd root has no restriction. Answer: 4; −1.
5
Solve x² − 6x − 2√(x² − 6x) − 8 = 0 with z = √(x² − 6x).
z² − 2z − 8 = 0 ⇒ z = 4 (z ≥ 0; z = −2 is rejected). x² − 6x = 16 ⇒ x = 8 or −2.