# How do you rationalize the denominator and simplify sqrt(3x)/(sqrtx-sqrt3)?

Jul 5, 2017

See a solution process below:

#### Explanation:

To rationalize the denominator multiply the expression by this form of $1$:

$\frac{\sqrt{x} + \sqrt{3}}{\sqrt{x} + \sqrt{3}}$

$\frac{\sqrt{x} + \sqrt{3}}{\textcolor{red}{\sqrt{x}} + \textcolor{red}{\sqrt{3}}} \times \frac{\sqrt{3 x}}{\textcolor{b l u e}{\sqrt{x}} - \textcolor{b l u e}{\sqrt{3}}} \implies$

$\frac{\sqrt{3 x} \left(\sqrt{x} + \sqrt{3}\right)}{\left(\textcolor{red}{\sqrt{x}} \textcolor{b l u e}{\sqrt{x}}\right) - \left(\textcolor{red}{\sqrt{x}} \textcolor{b l u e}{\sqrt{3}}\right) + \left(\textcolor{red}{\sqrt{3}} \textcolor{b l u e}{\sqrt{x}}\right) - \left(\textcolor{red}{\sqrt{3}} \textcolor{b l u e}{\sqrt{3}}\right)} \implies$

$\frac{\sqrt{3 x} \left(\sqrt{x} + \sqrt{3}\right)}{x - \left(\textcolor{b l u e}{\sqrt{3}} \textcolor{red}{\sqrt{x}}\right) + \left(\textcolor{red}{\sqrt{3}} \textcolor{b l u e}{\sqrt{x}}\right) - 3} \implies$

$\frac{\sqrt{3 x} \left(\sqrt{x} + \sqrt{3}\right)}{x - 0 - 3} \implies$

$\frac{\sqrt{3 x} \left(\sqrt{x} + \sqrt{3}\right)}{x - 3} \implies$

Now, we can work on simplifying the numerator:

$\frac{\textcolor{red}{\sqrt{3 x}} \left(\sqrt{x} + \sqrt{3}\right)}{x - 3} \implies$

$\frac{\left(\textcolor{red}{\sqrt{3 x}} \times \sqrt{x}\right) + \left(\textcolor{red}{\sqrt{3 x}} \times \sqrt{3}\right)}{x - 3} \implies$

$\frac{\sqrt{3 {x}^{2}} + \sqrt{9 x}}{x - 3} \implies$

$\frac{\sqrt{3} x + 3 \sqrt{x}}{x - 3}$