# Rationalizing the Denominator

Usually in algebra classes it's considered good practice to give your final answer without any radical signs in the denominator. In most cases, this can be accomplished by multiplying by some form of $1$ .

Example 1:

Simplify.

$\frac{\sqrt{5}}{\sqrt{6}}$

Multiply by $1$ in the form of $\frac{\sqrt{6}}{\sqrt{6}}$ .

$\frac{\sqrt{5}}{\sqrt{6}}=\frac{\sqrt{5}}{\sqrt{6}}\cdot \frac{\sqrt{6}}{\sqrt{6}}$

Simplify.

$=\frac{\sqrt{30}}{6}$

In other cases, if the denominator is a binomial with a rational part and an irrational part, then you'll need to use the conjugate of the binomial.

Example 2:

Simplify.

$\frac{1}{4-3\sqrt{7}}$

Multiply both the numerator and denominator by the conjugate of the denominator.

$\frac{1}{4-3\sqrt{7}}=\frac{1}{4-3\sqrt{7}}\cdot \frac{4+3\sqrt{7}}{4+3\sqrt{7}}$

The denominator is now a difference of squares .

$=\frac{4+3\sqrt{7}}{{4}^{2}-{\left(3\sqrt{7}\right)}^{2}}$

Use the power of a product property in the denominator.

$=\frac{4+3\sqrt{7}}{{4}^{2}-9\cdot 7}$

$=\frac{4+3\sqrt{7}}{16-63}$

$=\frac{-4-3\sqrt{7}}{47}$