How do you solve sqrt(x+2)=x-4 and find any extraneous solutions?

Jun 2, 2016

Square both sides of the equation to get rid of the square root.

Explanation:

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

$x + 2 = {x}^{2} - 8 x + 16$

$0 = {x}^{2} - 9 x + 14$

$0 = \left(x - 7\right) \left(x - 2\right)$

$x = 7 \mathmr{and} 2$

Checking back in the original equation we find that only $x = 7$ works. Therefore, the solution set is $\left\{x = 7\right\}$.

Hopefully this helps!