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

Jun 29, 2016

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

$x = {x}^{2} - 4 x + 4$

$0 = {x}^{2} - 5 x + 4$

$0 = \left(x - 4\right) \left(x - 1\right)$

$x = 4 \mathmr{and} 1$

Checking in the original equation, we find that only $x = 4$ works.

The solution set is $\left\{4\right\}$

Hopefully this helps!