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

Aug 28, 2016

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

#### Explanation:

Given:

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

Square both sides (noting that this may introduce spurious solutions) to get:

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

Subtract ${x}^{2}$ from both sides to get:

$5 = 9 - 6 x$

Subtract $9$ from both sides to get:

$- 4 = - 6 x$

Divide both sides by $- 6$ to get:

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

This is a valid solution of the original equation:

$\sqrt{{\left(\frac{2}{3}\right)}^{2} + 5} = \sqrt{\frac{4}{9} + 5} = \sqrt{\frac{49}{9}} = \frac{7}{3} = \frac{9 - 2}{3} = 3 - \frac{2}{3}$

Aug 28, 2016

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

#### Explanation:

color(blue)(sqrt(x^2+5)=3-x

$\text{Square both sides to cancel out the radical sign-}$ sqrt

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

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

$\text{Use the formula }$ color(orange)((a-b)^2=a^2-2ab+b^2

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

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

$\text{Cancel}$ ${x}^{2}$ $\text{both sides}$

rarrcancel(x^2)+5=9-6x+cancel(x^2

$\rightarrow 5 = 9 - 6 x$

$\text{Rewrite the equation}$

$\rightarrow 6 x = 9 - 5$

$\rightarrow 6 x = 4$

color(green)(rArrx=4/6=2/3