# How do you solve x=sqrt(2x^2-8)+2 and identify any restrictions?

Apr 2, 2017

$x = 2$

Our restriction is that $x \ge 2$.

#### Explanation:

Subtract $2$ on both sides:

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

Square both sides:

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

Expand the left side:

${x}^{2} - 2 x - 2 x + 4 = 2 {x}^{2} - 8$

Combine like terms on the left side:

${x}^{2} - 4 x + 4 = 2 {x}^{2} - 8$

Move everything on the left side to the right side:

$0 = 2 {x}^{2} - 8 - \left({x}^{2} - 4 x + 4\right)$

Combine like terms on the right side:

${x}^{2} \textcolor{red}{+} 4 x - 12 = 0$

Factor:

$\left(x \textcolor{red}{+} 6\right) \left(x \textcolor{red}{-} 2\right) = 0$

Solve for x:

$x = 2 , - 6$

Let's check the solutions:

$x = \sqrt{2 {x}^{2} - 8} + 2$

$2 = \sqrt{2 {\left(2\right)}^{2} - 8} + 2$

$2 = \sqrt{2 \left(4\right) - 8} + 2$

$2 = \sqrt{8 - 8} + 2$

$2 = \sqrt{0} + 2$

2=2 color(white)(00)color(green)sqrt

$- 6 = \sqrt{2 {\left(- 6\right)}^{2} - 8} + 2$

$- 6 = \sqrt{2 \left(36\right) - 8} + 2$

$- 6 = \sqrt{72 - 8} + 2$

$- 6 = \sqrt{64} + 2$

$- 6 = 8 + 2$

$- 6 = 10 \textcolor{w h i t e}{00} \textcolor{red}{X}$

And! We have restrictions:

• we can't allow the term under the square root to be less than 0. What values of $x$ are allowed?

$2 {x}^{2} - 8 \ge 0$

$2 {x}^{2} \ge 8$

${x}^{2} \ge 4$

$x \ge 2 , \le - 2$

Which means that $- 2 < x < 2$ is disallowed.

• Also, from $x - 2 = \sqrt{2 {x}^{2} - 8}$, we can see that $x - 2 \ge 0 \implies x \ge 2$. It's this restriction that eliminates $x = - 6$ from the solutions.

Therefore the restriction is $x \ge 2$