# How do you solve sqrt(x-3) =1 and find any extraneous solutions?

Sep 3, 2016

$x = 4$

#### Explanation:

The first step is to $\textcolor{b l u e}{\text{square both sides}}$

$\Rightarrow {\left(\sqrt{x - 3}\right)}^{2} = {1}^{2}$

$\textcolor{\mathmr{and} a n \ge}{\text{Reminder }} \textcolor{red}{| \overline{\underline{\textcolor{w h i t e}{\frac{a}{a}} \textcolor{b l a c k}{\sqrt{a} \times \sqrt{a} = {\left(\sqrt{a}\right)}^{2} = a} \textcolor{w h i t e}{\frac{a}{a}} |}}}$

$\Rightarrow x - 3 = 1 \Rightarrow x = 1 + 3 = 4$

There are no extraneous solutions.