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Simplify the following

$$\frac{\sqrt{3}}{\sqrt{2}(\sqrt{6} - \sqrt{3})}$$

Apparently the answer is $\frac{1}{2} (2 + \sqrt{2})$ but can't for the life of me see how to get it. Any help is massively appreciated.

Thanks

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  • $\begingroup$ Welcome to math.SE! Please consider taking the time to read the faq to familiarise yourself with some of our common practices. In addition, this page should give you a start at learning how to typeset mathematics here so that your posts say what you want them to, and also look good. $\endgroup$ – Najib Idrissi Oct 8 '14 at 7:43
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$\dfrac{\sqrt 3}{\sqrt 2 (\sqrt 6 - \sqrt 3)} = \dfrac{\sqrt 3}{\sqrt 2 \sqrt 3(\sqrt2 - 1)} = \dfrac{1}{\sqrt 2 (\sqrt 2 - 1)} = \dfrac{1}{(2 - \sqrt 2)} = \dfrac{(2 + \sqrt 2)}{(2 - \sqrt 2)(2 + \sqrt 2)} = \dfrac{2 + \sqrt 2}{2}$

The reason we write it this way is that it is bad practice to have roots at the denominator. To get rid of $a+b \sqrt c$ at the denominator, simply multiply the numerator and the denominator by $a-b \sqrt c$.

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  • $\begingroup$ Ok that makes sense. I was really just missing your first step of factorising the bracket in the denominator. Anyway nice solution and thanks for your help. $\endgroup$ – Jon Oct 8 '14 at 8:08
  • $\begingroup$ Why is it "bad practice to have roots in the denominator"? $\endgroup$ – Dan Fox Jan 31 at 8:32
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$\frac{\sqrt{3}}{\sqrt{2}(\sqrt{6}-\sqrt{3})}=\frac{\sqrt{1}}{\sqrt{2}(\sqrt{2}-\sqrt{1})}=\frac{\sqrt{2}+\sqrt{1}}{\sqrt{2}}=\frac12(2+\sqrt{2}).$

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  • $\begingroup$ Ok Paul I can see how you got there. Thanks for your response. $\endgroup$ – Jon Oct 8 '14 at 8:22

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