# How do you solve sqrt(2x+2)= sqrt( x^2-6)?

Apr 17, 2016

$x = 4$

#### Explanation:

Since you're dealing with square roots, it's always a good idea to start by writing down the valid solution intervals for $x$.

When working with real numbers, you can only take the square root of positive numbers, which means that you'll need

$2 x + 2 \ge 0$

$2 x \ge - 2 \implies x \ge - 1$

and

${x}^{2} - 6 \ge 0$

$\left(x - \sqrt{6}\right) \left(x + \sqrt{6}\right) \ge 0$

This takes place when you have

$x \in \left(- \infty , - \sqrt{6}\right] \cup \left[\sqrt{6} , + \infty\right)$

Combine these two restrictions, $x \ge - 1$ and $x \in \left(- \infty , - \sqrt{6}\right] \cup \left[\sqrt{6} , + \infty\right)$ to get

$\textcolor{p u r p \le}{| \overline{\underline{\textcolor{w h i t e}{\frac{a}{a}} \textcolor{b l a c k}{x \in \left[\sqrt{6} , + \infty\right)} \textcolor{w h i t e}{\frac{a}{a}} |}}}$

In order for a value of $x$ to be a valid solution to the original equation, you need it to come from that interval.

Now, square both sides of the equation to get rid of the square roots

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

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

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

You can calculate the two solutions by using the quadratic formula

$\textcolor{b l u e}{| \overline{\underline{\textcolor{w h i t e}{\frac{a}{a}} {x}_{1 , 2} = \frac{- b \pm \sqrt{{b}^{2} - 4 \cdot a \cdot c}}{2 \cdot a} \textcolor{w h i t e}{\frac{a}{a}} |}}}$

Here $a$, $b$, an $c$ are the coefficients of the quadratic equation $a {x}^{2} + b x + c = 0$.

${x}_{1 , 2} = \frac{- \left(- 2\right) \pm \sqrt{{\left(- 2\right)}^{2} - 4 \cdot 2 \cdot \left(- 8\right)}}{2 \cdot 1}$

${x}_{1 , 2} = \frac{2 \pm \sqrt{36}}{2}$

${x}_{1 , 2} = \frac{2 \pm 6}{2} \implies \left\{\begin{matrix}{x}_{1} = \frac{2 - 6}{2} = \textcolor{red}{\cancel{\textcolor{b l a c k}{- 2}}} \\ {x}_{2} = \frac{2 + 6}{2} = 4 \textcolor{w h i t e}{a} \textcolor{g r e e n}{\sqrt{}}\end{matrix}\right.$

Now, only one of these two values will be a valid solution to the original equation. Since

$x = - 2 \text{ } \notin \left[\sqrt{6} , + \infty\right)$

you can say that $x = - 2$ will be an extraneous solution. The only valid solution to the original equation will thus be

$\textcolor{g r e e n}{| \overline{\underline{\textcolor{w h i t e}{\frac{a}{a}} \textcolor{b l a c k}{x = 4} \textcolor{w h i t e}{\frac{a}{a}} |}}}$

Do a quick double-check to make sure that the calculations are correct

$\sqrt{2 \cdot 4 + 2} = \sqrt{{4}^{2} - 6}$

$\sqrt{10} = \sqrt{10} \textcolor{w h i t e}{a} \textcolor{g r e e n}{\sqrt{}}$