# Question #7eb8c

May 7, 2015

$\sqrt{2 x + 1} - \sqrt{x + 1} = 2$

Square both sides:
$\left(2 x + 1\right) - 2 \sqrt{\left(2 x + 1\right) \left(x + 1\right)} + \left(x + 1\right) = 4$

$3 x + 2 - 2 \sqrt{\left(2 x + 1\right) \left(x + 1\right)} = 4$

$\sqrt{\left(2 x + 1\right) \left(x + 1\right)} = \frac{3 x - 2}{2}$

Squaring both sides again:
$\left(2 x + 1\right) \left(x + 1\right) = \frac{9 {x}^{2} - 12 x + 4}{4}$

$4 \left(2 {x}^{2} + 3 x + 1\right) = 9 {x}^{2} - 12 x + 4$

$8 {x}^{2} + 12 x + 4 = 9 {x}^{2} - 12 x + 4$

${x}^{2} - 24 x = 0$

$x = 0$ or $x = 24$

Testing $x = 0$ in $\sqrt{2 x + 1} - \sqrt{x + 1} \ne 0$
so $x = 0$ is an extraneous solution.

Testing $x = 24$ in $\sqrt{2 x + 1} - \sqrt{x + 1}$
$= \sqrt{49} - \sqrt{25}$
$= 9 - 5$
$= 2$

$x = 24$ is the valid solution.