# How do you solve and check for extraneous solutions in -z=sqrt( -z+6)?

Aug 4, 2015

Valid solution $z = - 3$
(plus extraneous solution $z = 2$)

#### Explanation:

Given $- z = \sqrt{- z + 6}$
Square both sides
$\textcolor{w h i t e}{\text{XXXX}}$${z}^{2} = - z + 6$
Re-arrange into standard form
$\textcolor{w h i t e}{\text{XXXX}}$${z}^{2} + z - 6 = 0$
Factor
$\textcolor{w h i t e}{\text{XXXX}}$$\left(z + 3\right) \left(z - 2\right) = 0$

$\Rightarrow$
$\textcolor{w h i t e}{\text{XXXX}}$$z = - 3 \mathmr{and} z = + 2$

Checking
$\textcolor{w h i t e}{\text{XXXX}}$with $z = - 3$
$\textcolor{w h i t e}{\text{XXXX}}$$\textcolor{w h i t e}{\text{XXXX}}$$- z = \sqrt{- z + 6}$
$\textcolor{w h i t e}{\text{XXXX}}$becomes
$\textcolor{w h i t e}{\text{XXXX}}$$\textcolor{w h i t e}{\text{XXXX}}$$3 = \sqrt{3 + 6} = 3$
$\textcolor{w h i t e}{\text{XXXX}}$Valid

$\textcolor{w h i t e}{\text{XXXX}}$with $z = 2$
$\textcolor{w h i t e}{\text{XXXX}}$$\textcolor{w h i t e}{\text{XXXX}}$$- z = \sqrt{- z + 6}$
$\textcolor{w h i t e}{\text{XXXX}}$becomes
$\textcolor{w h i t e}{\text{XXXX}}$$\textcolor{w h i t e}{\text{XXXX}}$-2 ?= sqrt(-2+6) = 2
$\textcolor{w h i t e}{\text{XXXX}}$Extraneous