Question

What is the solution to the equation squareroot 4t+5 =3- squareroot t+5
x = –1, x = 11
x = –1
x = 11
no solution

Answer 1

Solve sqrt(4t+5) = 3 − sqrt(t+5).

Domain: 4t+5 ≥ 0 ⇒ t ≥ −5/4, and 3 − sqrt(t+5) ≥ 0 ⇒ sqrt(t+5) ≤ 3 ⇒ t ≤ 4. So −1.25 ≤ t ≤ 4.

Square both sides:
4t+5 = 9 − 6√(t+5) + (t+5) ⇒ 3t − 9 = −6√(t+5) ⇒ 3 − t = 2√(t+5).

Square again:
(3 − t)^2 = 4(t+5) ⇒ t^2 − 10t − 11 = 0 ⇒ t = 11 or t = −1.

Check in the original equation: t = −1 gives 1 = 1 (works). t = 11 gives 7 = −1 (false, extraneous).

Answer: t = −1 (or x = −1).