# Rationalizing fractions with multiple radicals

I am having trouble rationalizing $\frac{2}{\sqrt{2-\sqrt{2}}}$

I have tried multiplying the fraction by $\sqrt{2 + \sqrt{2}}$ and got $\frac{2\sqrt{2+\sqrt{2}}}{\sqrt{2}}$ I am not sure if that is correct or not but I then multiplied by $\sqrt{2}$ but got stuck...

Assuming that I did the problem correctly so far, how would I multiply $2\sqrt{2+\sqrt{2}}$ with $\sqrt{2}$ ?

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But $\sqrt{2 - \sqrt 2}\sqrt{2 + \sqrt 2} = \sqrt{4 - 2} = \sqrt 2$. How did you get $4- \sqrt 2$? And $2 \sqrt{2 + \sqrt 2}\sqrt 2 = 2 \sqrt{4 + 2\sqrt 2}$. –  martini Nov 5 '12 at 22:35
Could you explain how you multiplied $\sqrt{2}$ with $2\sqrt{2+\sqrt{2}}$? –  Kot Nov 5 '12 at 22:36
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## 1 Answer

Instead of multiplying top and bottom by $\sqrt{2}$, note that $2=\left(\sqrt{2}\right)^2$, so that $$\frac{2\sqrt{2+\sqrt{2}}}{\sqrt{2}}=\sqrt{2}\sqrt{2+\sqrt{2}}=\sqrt{4+2\sqrt{2}}.$$

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Does this mean that 'rationalizing' means getting rid of radicals only in the denominator, but not in the whole expression? –  Alex Nov 6 '12 at 0:59
Yes. We talk about "rationalizing the denominator" of a fraction with radicals in the denominator. Having a rational (integer, really) denominator makes arithmetic easier. We can't necessarily get rid of all the radicals, though, as we saw in this case. If we're lucky, we can get rid of all of the radicals--for an easy example, consider $\frac{\sqrt{8}}{\sqrt{2}}$--but most of the time we'll still have some hanging around. –  Cameron Buie Nov 6 '12 at 1:07
Ok that's cool. –  Alex Nov 6 '12 at 1:10
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