[SOLVED] Fractions with surds in the denominator

**waven** · March 16th 2009, 08:06 PM

Simply by rationalising the denominator

$\frac {\sqrt 3}<br /> {\sqrt 6}$

thanks!

**mathaddict** · March 16th 2009, 08:26 PM

Originally Posted by waven

Simply by rationalising the denominator

$\frac {\sqrt 3}<br /> {\sqrt 6}$

thanks!

multiply it by $\frac{\sqrt{6}}{\sqrt{6}}$

**waven** · March 16th 2009, 08:35 PM

Originally Posted by mathaddict

multiply it by $\frac{\sqrt{6}}{\sqrt{6}}$

$\frac {\sqrt 18} 6$ what do i do next?

btw the answer is $\frac {\sqrt 2} {2}$

**Prove It** · March 16th 2009, 10:11 PM

Originally Posted by waven

$\frac {\sqrt 18} 6$ what do i do next?

btw the answer is $\frac {\sqrt 2} {2}$

Notice that $\sqrt{18} = \sqrt{9 \times 2} = \sqrt{9} \times \sqrt{2} = 3\sqrt{2}$ .

So $\frac{\sqrt{18}}{6} = \frac{3\sqrt{2}}{6} = \frac{\sqrt{2}}{2}$ .

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