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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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Do you know that $(a+b)(a-b) = a^2 - b^2$? – user38268 Dec 2 '12 at 23:10
You can clear the denominator by multiplying top and bottom by the same $\sqrt 5 + 2.$ – Will Jagy Dec 2 '12 at 23:10
@WillJagy I did that and did not get a correct answer. – user138246 Dec 2 '12 at 23:12
$\frac{\sqrt{10}}{\sqrt{5} - 2} = \frac{\sqrt{10}(\sqrt{5} + 2)}{(\sqrt{5}-2)(\sqrt{5} + 2)} = ....$ – user38268 Dec 2 '12 at 23:16
@Debashish: Please do not use \dfrac in titles. – robjohn Jun 19 '14 at 10:37
up vote 3 down vote accepted

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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Wouldn't the conjugate be $-\sqrt{5} + 2$? – user138246 Dec 2 '12 at 23:45
@Jordan Yes, you are right! I have edited the answer. I multiplied by the symmetric of the conjugate of the denominator instead. The final result is the same. – Américo Tavares Dec 2 '12 at 23:47
So would the symmetric, conjugate and conjugate of the symmetric all work? – user138246 Dec 2 '12 at 23:50
@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. – Américo Tavares Dec 2 '12 at 23:54

$$ \frac{\sqrt{10}}{\sqrt5-2}=\frac{\sqrt{10}}{\sqrt5-2}\,\frac{\sqrt5+2}{\sqrt5+2}=\frac{\sqrt{10}(\sqrt5+2)}{5-4}=\sqrt{10}(\sqrt5+2) $$

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That last step is just magic to me, what happened? – user138246 Dec 2 '12 at 23:18
You mean $5-4=1$? – Martin Argerami Dec 2 '12 at 23:20
@Jordan $\frac{ \sqrt{10}(\sqrt{5} + 2)}{5 - 4} = \frac{\sqrt{10}(\sqrt{5} + 2)}{1} = \sqrt{10}(\sqrt{5} + 2)$ – user38268 Dec 2 '12 at 23:24
@MartinArgerami No, everything leading up to that you just magicked out of nowhere without any explanation. If you want to see what a real explanation looks like refer to Americo's answer. The one I accepted. – user138246 Dec 3 '12 at 0:06
@Jordan: there is no reason to be nasty. If you look at what Martin is doing, you will see that he is exploiting the fact that $(x+y)(x-y)=x^2-y^2$ to get rid of a difference (or sum) of square roots ($2=\sqrt{4}$) in the denominator. While it is true that Américo's answer explains this more explicitly, discovering this by studying why this answer works can be just as educational. – robjohn Dec 3 '12 at 0:33

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