# How do you solve 2sqrt(x-11)-8=4 and check your solution?

May 4, 2017

See a solution process below:

#### Explanation:

First, add $\textcolor{red}{8}$ to each side of the equation to isolate the radical term while keeping the equation balanced:

$2 \sqrt{x - 11} - 8 + \textcolor{red}{8} = 4 + \textcolor{red}{8}$

$2 \sqrt{x - 11} - 0 = 12$

$2 \sqrt{x - 11} = 12$

Next, divide each side of the equation by $\textcolor{red}{2}$ to isolate the radical while keeping the equation balanced:

$\frac{2 \sqrt{x - 11}}{\textcolor{red}{2}} = \frac{12}{\textcolor{red}{2}}$

$\frac{\textcolor{red}{\cancel{\textcolor{b l a c k}{2}}} \sqrt{x - 11}}{\cancel{\textcolor{red}{2}}} = 6$

$\sqrt{x - 11} = 6$

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

${\left(\sqrt{x - 11}\right)}^{2} = {6}^{2}$

$x - 11 = 36$

Now, add $\textcolor{red}{11}$ to each side of the equation to solve for $x$ while keeping the equation balanced:

$x - 11 + \textcolor{red}{11} = 36 + \textcolor{red}{11}$

$x - 0 = 47$

$x = 47$

To check the solution substitute $\textcolor{red}{47}$ for $\textcolor{red}{x}$ in the original equation and calculate the left side of the equation to ensure it equals $4$

$2 \sqrt{\textcolor{red}{x} - 11} - 8 = 4$ becomes:

$2 \sqrt{\textcolor{red}{47} - 11} - 8 = 4$

$2 \sqrt{36} - 8 = 4$

$\left(2 \times \pm 6\right) - 8 = 4$

$\pm 12 - 8 = 4$

$4 = 4$

Or

$- 20 \ne 4$

The solution of $- 20$ is extraneous.