# How do you solve sqrt(9x+10)=x and find any extraneous solutions?

May 27, 2016

$x = 10$ or $x = - 1$
(no extraneous solutions)

#### Explanation:

Given
$\textcolor{w h i t e}{\text{XXX}} \sqrt{9 x + 10} = x$

Square both sides (this where an extraneous solution might be introduced)
$\textcolor{w h i t e}{\text{XXX}} 9 x + 10 = {x}^{2}$
Rearrange as a quadratic in standard form
$\textcolor{w h i t e}{\text{XXX}} {x}^{2} - 9 x - 10 = 0$
Factor
$\textcolor{w h i t e}{\text{XXX}} \left(x - 10\right) \left(x + 1\right) = 0$
Either
$\textcolor{w h i t e}{\text{XXX")(x-10)=0color(white)("XX")rarrcolor(white)("XX}} x = 10$
or
$\textcolor{w h i t e}{\text{XXX")(x+1)=0color(white)("XX")rarrcolor(white)("XX}} x = - 1$

Checking against original equation:
$\textcolor{w h i t e}{\text{XXX}}$If $x = 10$
$\textcolor{w h i t e}{\text{XXXXXXX}} \sqrt{9 x + 10} = \sqrt{9 \left(10\right) + 10} = \sqrt{100} = 10 = x$
$\textcolor{w h i t e}{\text{XXX}}$This solution is valid.

$\textcolor{w h i t e}{\text{XXX}}$If $x = - 1$
$\textcolor{w h i t e}{\text{XXXXXX}} \sqrt{9 x + 10} = \sqrt{9 \left(- 1\right) + 10} = \sqrt{1} = 1 = x$
$\textcolor{w h i t e}{\text{XXX}}$This solution is valid