# How do you solve sqrt(4x+5)=x?

Aug 2, 2016

${\left(\sqrt{4 x + 5}\right)}^{2} = {x}^{2}$

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

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

$0 = \left(x - 5\right) \left(x + 1\right)$

$x = 5 \mathmr{and} - 1$

However, checking in the original equation, we find that only $x = 5$ works. Hence, the solution set is $\left\{5\right\}$.

Hopefully this helps!