# How do you solve sqrt(x^2+2)-5=0?

Aug 20, 2017

$x = \sqrt{23} , x = - \sqrt{23}$

#### Explanation:

$\sqrt{{x}^{2} + 2} - 5 = 0$

$\sqrt{{x}^{2} + 2} = 5$

Square each side:
${x}^{2} + 2 = 25$

Subtract 2 from each side:
${x}^{2} = 23$

$\textcolor{b l u e}{x = \pm \sqrt{23}}$

Aug 20, 2017

See a solution process below:

#### Explanation:

First, add $\textcolor{red}{5}$ to each side of the equation to isolate the radical while keeping the equation balanced:

$\sqrt{{x}^{2} + 2} - 5 + \textcolor{red}{5} = 0 + \textcolor{red}{5}$

$\sqrt{{x}^{2} + 2} - 0 = 5$

$\sqrt{{x}^{2} + 2} = 5$

Next, square each side of the equation to eliminate the radical while keeping the equation balanced:

${\left(\sqrt{{x}^{2} + 2}\right)}^{2} = {5}^{2}$

${x}^{2} + 2 = 25$

Then, subtract $\textcolor{red}{2}$ from each side of the equation to isolate ${x}^{2}$ while keeping the equation balanced:

${x}^{2} + 2 - \textcolor{red}{2} = 25 - \textcolor{red}{2}$

${x}^{2} + 0 = 23$

${x}^{2} = 23$

Now, take the square root of each side of the equation to solve for $x$ while keeping the equation balanced. Remember, the square root of a number produces both a positive and a negative result:

$\sqrt{{x}^{2}} = \pm \sqrt{23}$

$x = \pm \sqrt{23}$