# How do you solve sqrt(x+2) = x-4?

May 10, 2016

$x = 7$

#### Explanation:

First square both sides of the equation, noting that this may introduce spurious solutions:

$x + 2 = {\left(x - 4\right)}^{2} = {x}^{2} - 8 x + 16$

Subtract $x + 2$ from both ends to get:

$0 = {x}^{2} - 9 x + 14 = \left(x - 2\right) \left(x - 7\right)$

So $x = 2$ or $x = 7$

If $x = 2$ then $\sqrt{x + 2} = \sqrt{4} = 2 \ne - 2 = 2 - 4 = x - 4$

If $x = 7$ then $\sqrt{x + 2} = \sqrt{9} = 3 = 7 - 4 = x - 4$

So $x = 7$ is the only valid solution of the original equation.