# How do you rationalize the denominator 7/(sqrt3-sqrt2) ?

May 26, 2015

You can multiply and divide your expression by $\sqrt{3} + \sqrt{2}$ to get:
$\frac{7}{\sqrt{3} - \sqrt{2}} \cdot \frac{\sqrt{3} + \sqrt{2}}{\sqrt{3} + \sqrt{2}} =$

in the denominator you have a notable:
$\left(a + b\right) \left(a - b\right) = {a}^{2} - {b}^{2}$

So you get:
$\frac{7 \left(\sqrt{3} + \sqrt{2}\right)}{3 - 2} = 7 \left(\sqrt{3} + \sqrt{2}\right)$

May 26, 2015

Multiply numerator and denominator by the conjugate $\left(\sqrt{3} + \sqrt{2}\right)$:

$\frac{7}{\sqrt{3} - \sqrt{2}}$

$= \frac{7 \left(\sqrt{3} + \sqrt{2}\right)}{\left(\sqrt{3} - \sqrt{2}\right) \left(\sqrt{3} + \sqrt{2}\right)}$

$= \frac{7 \left(\sqrt{3} + \sqrt{2}\right)}{{\sqrt{3}}^{2} - {\sqrt{2}}^{2}}$

$= \frac{7 \left(\sqrt{3} + \sqrt{2}\right)}{3 - 2}$

$= 7 \left(\sqrt{3} + \sqrt{2}\right)$