How do you solve and check for extraneous solutions in sqrt(x+3) =x-3?

Jul 31, 2015

Square both sides, solve the resulting quadratic equation and discard the extraneous solution to find: $x = 6$

Explanation:

First note that we require $x \ge 3$ since the left hand side is non-negative.

Now square both sides to get:
$x + 3 = {x}^{2} - 6 x + 9$

Subtract $x + 3$ from both sides to get:

$0 = {x}^{2} - 7 x + 6 = \left(x - 1\right) \left(x - 6\right)$

This has two solutions: $x = 1$ and $x = 6$, but $x = 1$ is not a solution of the original equation since we require $x \ge 3$.

So the only solution to the original equation is $x = 6$.