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.
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.
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] $\frac{1}{a+\sqrt{b}}=\frac{a-\sqrt{b}}{\left( a+\sqrt{b}\right) \left( a-\sqrt{b}\right) }=\frac{a-\sqrt{b}}{a^{2}-b}.$
$ \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) $