# How do you rationalize the denominator and simplify sqrtx/(sqrtx+sqrt3)?

Jun 30, 2016

$\frac{\sqrt{x}}{\sqrt{x} + \sqrt{3}} = \textcolor{b l u e}{\frac{x - \sqrt{3 x}}{x - 3}}$

#### Explanation:

For the general expression: $\left(a + b\right)$
the conjugate is $\left(a - b\right)$
and the product of an expression times its conjugate is
$\textcolor{w h i t e}{\text{XXX}} \left(a + b\right) \cdot \left(a - b\right) = {a}^{2} - {b}^{2}$

If $a$ and/or $b$ are square roots this lets us get rid of the square roots.

For the given example $\frac{\sqrt{x}}{\sqrt{x} + \sqrt{3}}$

multiplying both the numerator and denominator by the conjugate of $\left(\sqrt{x} + \sqrt{3}\right)$ lets us remove the square roots from the denominator.

$\textcolor{w h i t e}{\text{XXX}} \frac{\sqrt{x}}{\sqrt{x} + \sqrt{3}} \times \frac{\sqrt{x} - \sqrt{3}}{\sqrt{x} - \sqrt{3}}$

$\textcolor{w h i t e}{\text{XXX}} = \frac{\sqrt{x} \cdot \left(\sqrt{x} - \sqrt{3}\right)}{x - 3}$

$\textcolor{w h i t e}{\text{XXX}} = \frac{x - \sqrt{3 x}}{x - 3}$