# How do you solve x+3=sqrt(18-2x^2) algebraically?

Dec 26, 2016

$\left\{- 3\right\}$

#### Explanation:

Square both sides.

${\left(x + 3\right)}^{2} = {\left(\sqrt{18 - 2 {x}^{2}}\right)}^{2}$

${x}^{2} + 6 x + 9 = 18 - 2 {x}^{2}$

$3 {x}^{2} + 6 x - 9 = 0$

$3 \left({x}^{2} + 2 x - 3\right) = 0$

${x}^{2} + 2 x - 3 = 0$

$\left(x + 3\right) \left(x - 1\right) = 0$

$x = - 3 \mathmr{and} 1$

Checking in the original equation, you will find that only $x = - 3$ will work.

Hopefully this helps!