# How do you rationalize the denominator sqrt6/(sqrt13+3)?

Jun 2, 2015

Multiply both numerator (top) and denominator (bottom) by the conjugate $\left(\sqrt{13} - 3\right)$ ...

$\frac{\sqrt{6}}{\sqrt{13} + 3}$

$= \frac{\sqrt{6}}{\sqrt{13} + 3} \cdot \frac{\sqrt{13} - 3}{\sqrt{13} - 3}$

$= \frac{\sqrt{6} \cdot \left(\sqrt{13} - 3\right)}{\left(\sqrt{13} + 3\right) \left(\sqrt{13} - 3\right)}$

$= \frac{\sqrt{6} \cdot \left(\sqrt{13} - 3\right)}{13 - 9}$

$= \frac{\sqrt{6} \cdot \left(\sqrt{13} - 3\right)}{4}$