Question

What is the solution of StartRoot x + 12 EndRoot = x?
x = 4
x = –3 or x = 4
A or B

Answer 1

Solve √(x + 12) = x.

Domain: x + 12 ≥ 0 and √(...) ≥ 0 ⇒ x ≥ 0.

Square both sides: x + 12 = x^2 ⇒ x^2 − x − 12 = 0 ⇒ (x − 4)(x + 3) = 0 ⇒ x = 4 or x = −3.

Check domain: x = −3 is invalid (negative), x = 4 is valid: √(4+12)=√16=4.

So the solution is x = 4 (choice A).