How do you solve sqrt(2X-1) = 5 - sqrt(X-1) and find any extraneous solutions?

Mar 14, 2017

$\textcolor{g r e e n}{x = 5}$
$\textcolor{w h i t e}{\text{XXX}}$extraneous solution at $x = 145$

Explanation:

Given
$\textcolor{w h i t e}{\text{XXX}} \sqrt{2 x - 1} + \sqrt{x - 1} = 5$

Squaring both sides
$\textcolor{w h i t e}{\text{XXX}} 2 x - 1 + 2 \sqrt{2 x - 1} \sqrt{x - 1} + x - 1 = 25$

Simplifying
$\textcolor{w h i t e}{\text{XXX}} 3 x - 2 + 2 \sqrt{2 {x}^{2} - 3 x + 1} = 25$

$\textcolor{w h i t e}{\text{XXX}} 2 \sqrt{2 {x}^{2} - 3 x + 1} = 27 - 3 x$

Squaring both sides again
$\textcolor{w h i t e}{\text{XXX}} 8 {x}^{2} - 12 x + 4 = 729 - 162 x + {x}^{2}$

$\textcolor{w h i t e}{\text{XXX}} {x}^{2} - 150 x + 725$

Factoring (or use the quadratic formula if the factors are not apparent)
$\textcolor{w h i t e}{\text{XXX}} \left(x - 5\right) \left(x - 145\right) = 0$

Giving possible solutions:
$\textcolor{w h i t e}{\text{XXX")x=5color(white)("XX")andcolor(white)("XX}} x = 145$

Substituting back into the original equation:
[1]$\textcolor{w h i t e}{\text{XXX}} x = 5$
$\textcolor{w h i t e}{\text{XXXXX}} L S$
$\textcolor{w h i t e}{\text{XXXXXXX}} = \sqrt{2 \cdot 5 - 1}$
$\textcolor{w h i t e}{\text{XXXXXXX}} = \sqrt{9}$
$\textcolor{w h i t e}{\text{XXXXXXX}} = 3$
$\textcolor{w h i t e}{\text{XXXXX}} R S$
$\textcolor{w h i t e}{\text{XXXXXXX}} = 5 - \sqrt{5 - 1}$
$\textcolor{w h i t e}{\text{XXXXXXX}} = 5 - 2$
$\textcolor{w h i t e}{\text{XXXXXXX}} = 3$
$\textcolor{w h i t e}{\text{XXX")LS=RScolor(white)("XXX}}$valid

[2]$\textcolor{w h i t e}{\text{XXX}} x = 145$
$\textcolor{w h i t e}{\text{XXX}} L S$
$\textcolor{w h i t e}{\text{XXXXX}} = \sqrt{2 \cdot 145 - 1}$
$\textcolor{w h i t e}{\text{XXXXXXX}} = \sqrt{289}$
$\textcolor{w h i t e}{\text{XXXXXXX}} = 17$
$\textcolor{w h i t e}{\text{XXX}} R S$
$\textcolor{w h i t e}{\text{XXXXXXX}} = \sqrt{145 - 1}$
$\textcolor{w h i t e}{\text{XXXXXXX}} = \sqrt{144}$
$\textcolor{w h i t e}{\text{XXXXXXX}} = 12$
$\textcolor{w h i t e}{\text{XXX")LS!=RScolor(white)("XXX}}$extraneous