Question

What is the solution to the equation \sqrt{x^{2}+2x-25}=\sqrt{x+5} ? Assume the range is all real numbers.

x = –6
x = 5
x = 6, x = –5
x = –6, x = 5

Answer 1

Square both sides: x^2+2x-25 = x+5 → x^2+x-30=0 → (x+6)(x-5)=0, so x=-6 or x=5.
But domain requires x+5 ≥ 0 ⇒ x ≥ -5, so x=-6 is extraneous.
Solution: x = 5.