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

Jun 10, 2017

See a solution process below:

#### Explanation:

First, add $\textcolor{red}{6}$ to each side of the equation to isolate the $h$ term while keeping the equation balanced:

$4 + \textcolor{red}{6} = \sqrt{h} - 6 + \textcolor{red}{6}$

$10 = \sqrt{h} - 0$

$10 = \sqrt{h}$

Now, square each side of the equation to solve for $h$ while keeping the equation balanced:

${10}^{2} = {\left(\sqrt{h}\right)}^{2}$

$100 = h$

$h = 100$

There are no extraneous solutions for $h$.

However, if we substitute $100$ for $h$ we get:

$4 = \pm \sqrt{100} - 6$

$4 = \pm 10 - 6$

The solution: $4 = + 10 - 6$ is valid.

The solution: $4 = - 10 - 6$ is not valid.

But there is only one solution for $h$ which is $100$.