# How do you solve x/(3 sqrt (2x-3)) - 1/ (sqrt (2x-3))=1/3 and find any extraneous solutions?

Apr 9, 2018

$x = 6$

#### Explanation:

$\frac{x}{3 \sqrt{2 x - 3}} - \frac{1}{\sqrt{2 x - 3}} = \frac{1}{3}$

$\frac{x - 3}{3 \sqrt{2 x - 3}} = \frac{1}{3}$

$x - 3 = \sqrt{2 x - 3}$

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

${x}^{2} - 6 x + 9 = 2 x - 3$

${x}^{2} - 8 x + 12 = 0$

$\left(x - 2\right) \cdot \left(x - 6\right) = 0$

So ${x}_{1} = 2$, ${x}_{2} = 6$. But $x = 2$ is extraneous solution. Hence $x = 6$ is solution of this equation.