# How do you solve  sqrt[x+2]=sqrt[4-x] and find any extraneous solutions?

Oct 10, 2017

$x = 1$

#### Explanation:

$\text{to access the contents of the radicals}$

$\textcolor{b l u e}{\text{square both sides}}$

•color(white)(x)sqrtaxxsqrta=(sqrta)^2=a

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

$\Rightarrow x + 2 = 4 - x$

$\text{add x to both sides}$

$x + x + 2 = 4 \cancel{- x} \cancel{+ x}$

$\Rightarrow 2 x + 2 = 4$

$\text{subtract 2 from both sides}$

$2 x \cancel{+ 2} \cancel{- 2} = 4 - 2$

$\Rightarrow 2 x = 2$

$\text{divide both sides by 2}$

$\frac{\cancel{2} x}{\cancel{2}} = \frac{2}{2}$

$\Rightarrow x = 1$

$\textcolor{b l u e}{\text{As a check}}$

Substitute this value into the equation and if both sides are equal then it is the solution.

$\text{left side } = \sqrt{1 + 2} = \sqrt{3}$

$\text{right side } = \sqrt{4 - 1} = \sqrt{3}$

$\Rightarrow x = 1 \text{ is the solution}$

There are no other solutions hence no extraneous solutions.