# How do you solve sqrt(3x-5)-sqrt(3x)=-1?

Aug 5, 2015

$x = 3$

#### Explanation:

The first thing you need to do is square both sides of the equation. This will leave you with only one radical term.

${\left(\sqrt{3 x - 5} - \sqrt{3 x}\right)}^{2} = {\left(- 1\right)}^{2}$

$\left[{\left(\sqrt{3 x - 5}\right)}^{2} - 2 \cdot \sqrt{\left(3 x - 5\right) \cdot 3 x} + {\left(\sqrt{3 x}\right)}^{2}\right] = 1$

$\left(3 x - 5 - 2 \sqrt{9 {x}^{2} - 15 x} + 3 x\right) = 1$

$6 x - 5 - 2 \sqrt{9 {x}^{2} - 15 x} = 1$

Isolate the remaining radical term on onse side of the equation, then square both sides again.

$- 2 \sqrt{9 {x}^{2} - 15 x} = 6 - 6 x$

This is equivalent to

$\sqrt{9 {x}^{2} - 15 x} = 3 x - 3$

${\left(\sqrt{9 {x}^{2} - 15 x}\right)}^{2} = {\left(3 x - 3\right)}^{2}$

$\textcolor{red}{\cancel{\textcolor{b l a c k}{9 {x}^{2}}}} - 15 x = \textcolor{red}{\cancel{\textcolor{b l a c k}{9 {x}^{2}}}} - 18 x + 9$

Rearrange to isolate $x$ on onse side of the equation

$18 x - 15 x = 9$

$3 x = 9 \implies x = \frac{9}{3} = \textcolor{g r e e n}{3}$