How do you rationalize the denominator and simplify $\frac{3}{\sqrt{6} - \sqrt{3}}$ $3/(sqrt6-sqrt3)$ ?

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How do you rationalize the denominator and simplify $\frac{3}{\sqrt{6} - \sqrt{3}}$ $3/(sqrt6-sqrt3)$ ?

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Answer 1 · 2015-04-01T02:12:51

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

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

$= \frac{3 \cdot \sqrt{3} \cdot (\sqrt{2} + 1)}{6 - 3}$ $= (3* sqrt(3) * (sqrt(2) + 1))/(6-3)$

$= \sqrt{3} \cdot (\sqrt{2} + 1)$ $= sqrt(3) * (sqrt(2) + 1)$

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How do you rationalize the denominator and simplify $\frac{3}{\sqrt{6} - \sqrt{3}}$ $3/(sqrt6-sqrt3)$ ?

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