How do you rationalize the denominator and simplify $\frac{6}{3 + \sqrt{3}}$ $6/(3+sqrt3)$ ?

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How do you rationalize the denominator and simplify $\frac{6}{3 + \sqrt{3}}$ $6/(3+sqrt3)$ ?

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Answer 1 · 2016-08-30T18:43:20

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

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 $a + b$ $a + b$ is $a - b$ $a -b$ , since $(a + b) (a - b) = a^{2} + a b - a b - b^{2} = a^{2} - b^{2}$ $(a + b)(a - b) = a^2 + ab - ab - b^2 = a^2 - b^2$ , or a difference of squares.

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

Let's start the rationalization process.

$= \frac{6}{3 + \sqrt{3}} \times \frac{3 - \sqrt{3}}{3 - \sqrt{3}}$ $=6/(3 + sqrt(3)) xx (3 - sqrt(3))/(3 - sqrt(3))$

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

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

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

$= 3 - \sqrt{3}$ $=3 - sqrt(3)$

Hopefully this helps!

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