# How do you solve 2+sqrt[x]=sqrt[x+4] and find any extraneous solutions?

##### 1 Answer
Jul 4, 2016

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

$4 + 4 \sqrt{x} + x = x + 4$

$4 \sqrt{x} = x - x + 4 - 4$

$4 \sqrt{x} = 0$

${\left(4 \sqrt{x}\right)}^{2} = {0}^{2}$

$16 x = 0$

$x = 0$

Checking in the original equation we find this solution works. The solution set is $\left\{0\right\}$.

Hopefully this helps!