# How do you rationalize (3sqrt3-2sqrt2)/ (3sqrt3+2sqrt2)?

May 17, 2015

Multiply both the numerator (top) and denominator (bottom) by the conjugate of the denominator:

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

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

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

$= \frac{27 - 12 \sqrt{6} + 8}{27 - 8}$

$= \frac{35 - 12 \sqrt{6}}{19}$

The reason you can eliminate the square roots from the denominator in this way is because $\left(a + b\right) \left(a - b\right) = {a}^{2} - {b}^{2}$ for any $a$ and $b$.