# How do you solve x=sqrt( -7x+8) and find any extraneous solutions?

Jul 16, 2016

${\left(x\right)}^{2} = {\left(\sqrt{- 7 x + 8}\right)}^{2}$

${x}^{2} = - 7 x + 8$

${x}^{2} + 7 x - 8 = 0$

$\left(x + 8\right) \left(x - 1\right) = 0$

$x = - 8 \mathmr{and} 1$

Checking in the original équation, we find only $x = 1$ works. Hence, $\left\{1\right\}$ is the solution set.

Hopefully this helps!