# How do you solve sqrt(8-2x) = x-4?

Oct 2, 2017

$x = 2$
$x = 4$

#### Explanation:

$\sqrt{8 - 2 x} = x - 4$

Square both sides to remove radical:

$8 - 2 x = {\left(x - 4\right)}^{2}$

Expand right hand side:

$8 - 2 x = {x}^{2} - 8 x + 16$

Collect like terms on one side of the equation:

${x}^{2} - 6 x + 8 = 0$

Factor ${x}^{2} - 6 x + 8 = 0$:

$\left(x - 2\right) \left(x - 4\right) = 0 \implies x = 2$ and $x = 4$