# How do you solve and check for extraneous solutions in sqrt(x^2+5)=3-x?

Aug 1, 2015

$\textcolor{red}{x = \frac{2}{3}}$
There are $\textcolor{red}{\text{no}}$ extraneous solutions.

#### Explanation:

$\sqrt{{x}^{2} + 5} = 3 - x$

Square each side.

${x}^{2} + 5 = {\left(3 - x\right)}^{2}$

Remove parentheses.

${x}^{2} + 5 = 9 - 6 x + {x}^{2}$

Move all terms in $x$ the left hand side.

${x}^{2} + 5 + 6 x - {x}^{2} = 9$

$5 + 6 x = 9$

$6 x = 9 - 5 = 4$

$x = \frac{4}{6}$

$x = \frac{2}{3}$

There are no extraneous solutions.

Check:

$\sqrt{{x}^{2} + 5} = 3 - x$

$\sqrt{{\left(\frac{2}{3}\right)}^{2} + 5} = 3 - \frac{2}{3}$

$\sqrt{\frac{4}{9} + 5} = \frac{9 - 2}{3}$

$\sqrt{\frac{45 + 4}{9}} = \frac{7}{3}$

$\sqrt{\frac{49}{9}} = \frac{7}{3}$

$\frac{7}{3} = \frac{7}{3}$