# How do you solve sqrt(10–x)+x=8 and find any extraneous solutions?

Sep 26, 2016

$\sqrt{10 - x} = 8 - x$

${\left(\sqrt{10 - x}\right)}^{2} = {\left(8 - x\right)}^{2}$

$10 - x = 64 - 16 x + {x}^{2}$

$0 = {x}^{2} - 15 x + 54$

$0 = \left(x - 9\right) \left(x - 6\right)$

$x = 9 \mathmr{and} 6$

Checking in the original equation, you will find that $x = 9$ is extraneous. However, $x = 6$ works. Hence, our solution set will be $\left\{6\right\}$.

Hopefully this helps!