How do you solve sqrt(2x+20) + 2 = x and find any extraneous solutions?

Feb 17, 2017

$\left[8\right\}$.

Explanation:

Always start by isolating the radical.

$\sqrt{2 x + 20} = x - 2$

Square both sides.

${\left(\sqrt{2 x + 20}\right)}^{2} = {\left(x - 2\right)}^{2}$

$2 x + 20 = {x}^{2} - 4 x + 4$

$0 = {x}^{2} - 6 x - 16$

$0 = \left(x - 8\right) \left(x + 2\right)$

$x = 8 \mathmr{and} - 2$

The last step to such problems is always to check your solutions within the original equation.

Check: $x = 8$

sqrt(2(8) +20) + 2 =^? 8

sqrt(36) + 2 =8 color(green)(√)

Check: $x = - 2$

sqrt(2(-2) + 20) + 2 = ^?-2

$\sqrt{16} + 2 \ne - 2$

Therefore, the solution set is $\left\{8\right\}$.

Hopefully this helps!