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I have the expression

$\frac {\sqrt{10}}{\sqrt{5} -2}$

I can't figure out what to do from here, I can't seem to pull any numbers out of either of the square roots so it appears that it must remain as is.

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    ok ! Thanks @robjohn2014-06-19

2 Answers 2

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To rationalize the denominator i. e. turn the denominator rational multiply both numerator and denominator by the conjugate of the denominator $-\sqrt{5}-2$ or its symmetric $\sqrt{5}+2$, expand both and simplify

$\begin{eqnarray*} \frac{\sqrt{10}}{\sqrt{5}-2} &=&\frac{\sqrt{10}\left( \sqrt{5}+2\right) }{ \left( \sqrt{5}-2\right) \left( \sqrt{5}+2\right) }=\frac{\sqrt{10}\sqrt{5}+\sqrt{10}\times 2}{\left( \sqrt{5}\right) ^{2}-2^{2}} \\ &=&\frac{\sqrt{50}+2\sqrt{10}}{5-4}=\frac{\sqrt{50}+2\sqrt{10}}{1}=\sqrt{50}% +2\sqrt{10}. \end{eqnarray*}. $

In general [Edited to correct] $\frac{1}{a+\sqrt{b}}=\frac{a-\sqrt{b}}{\left( a+\sqrt{b}\right) \left( a-\sqrt{b}\right) }=\frac{a-\sqrt{b}}{a^{2}-b}.$

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    @Jordan: Sorry, the conjugate is $-\sqrt{5}-2$ (the square root has the minus sign). The symmetric of the conjugate is $-(-\sqrt{5}-2)=\sqrt{5}+2$. Both conjugate and the symmetric of the conjugate work, but in this case the symmetric of the conjugate is easier! Try with the conjugate to convince yourself.2012-12-02
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$ \frac{\sqrt{10}}{\sqrt5-2}=\frac{\sqrt{10}}{\sqrt5-2}\,\frac{\sqrt5+2}{\sqrt5+2}=\frac{\sqrt{10}(\sqrt5+2)}{5-4}=\sqrt{10}(\sqrt5+2) $

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    @Jordan: It's kind of weird of you to say that, taking into account that before I posted my answer, you said to Will Jagy that you knew about the trick of multiplying numerator and denominator by $\sqrt5+2$.2012-12-03