# How do you rationalize the denominator and simplify 6/(3+sqrt3)?

Aug 30, 2016

The answer is $3 - \sqrt{3}$.

#### Explanation:

When rationalizing expressions where the denominators are binomials (two terms), we multiply the entire expression by the conjugate of the denominator.

The conjugate is meant to make a difference of squares when multiplied with its original expression. For example, the conjugate of $a + b$ is $a - b$, since $\left(a + b\right) \left(a - b\right) = {a}^{2} + a b - a b - {b}^{2} = {a}^{2} - {b}^{2}$, or a difference of squares.

The conjugate can always be found by switching the middle sign of the denominator. Hence, the conjugate of $3 + \sqrt{3}$ is $3 - \sqrt{3}$.

Let's start the rationalization process.

$= \frac{6}{3 + \sqrt{3}} \times \frac{3 - \sqrt{3}}{3 - \sqrt{3}}$

$= \frac{18 - 6 \sqrt{3}}{9 - 3 \sqrt{3} + 3 \sqrt{3} - \sqrt{9}}$

$= \frac{18 - 6 \sqrt{3}}{9 - 3}$

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

$= 3 - \sqrt{3}$

Hopefully this helps!