# How do you solve sqrt(6x-20)=4 and find any extraneous solutions?

Sep 10, 2017

See a solution process below:

#### Explanation:

First, square both sides of the equation to eliminate the square root while keeping the equation balanced:

${\left(\sqrt{6 x - 20}\right)}^{2} = {4}^{2}$

$6 x - 20 = 16$

Next, add $\textcolor{red}{20}$ to each side of the equation to isolate the $x$ term while keeping the equation balanced:

$6 x - 20 + \textcolor{red}{20} = 16 + \textcolor{red}{20}$

$6 x - 0 = 36$

$6 x = 36$

Now, divide each side of the equation by $\textcolor{red}{6}$ to solve for $x$ while keeping the equation balanced:

$\frac{6 x}{\textcolor{red}{6}} = \frac{36}{\textcolor{red}{6}}$

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

$x = 6$

If we substitute $\textcolor{red}{6}$ for $\textcolor{red}{x}$ in the original equation and evaluate the square root we will find the extraneous solutions. Remember, the square root of a number produces a positive and negative result:

$\sqrt{6 \textcolor{red}{x} - 20} = 4$ becomes:

$\pm \sqrt{\left(6 \cdot \textcolor{red}{6}\right) - 20} = 4$

Or

$- \sqrt{\left(6 \cdot \textcolor{red}{6}\right) - 20} = 4$ and $\sqrt{\left(6 \cdot \textcolor{red}{6}\right) - 20} = 4$

$- \sqrt{36 - 20} = 4$ and $\sqrt{36 - 20} = 4$

$- \sqrt{16} = 4$ and $\sqrt{16} = 4$

$- 4 \ne 4$ and $4 = 4$

$- \sqrt{16}$ is an extraneous solution.