Square root madness!

**usagi_killer** · April 26th 2009, 01:59 AM

Find $\sqrt{-\sqrt{3}+\sqrt{4+\sqrt{5}+\sqrt{17-4\sqrt{15}}}}$

**Moo** · April 26th 2009, 02:17 AM

Hello,

Originally Posted by usagi_killer

Find $\sqrt{-\sqrt{3}+\sqrt{4+\sqrt{5}+\sqrt{17-4\sqrt{15}}}}$

Note that $17-4 \sqrt{15}=5+2^2 \cdot 3-2 \cdot \sqrt{5}\cdot (2 \sqrt{3})=(\sqrt{5})^2+(2\sqrt{3})^2-2 \cdot \sqrt{5}\cdot (2 \sqrt{3})=(\sqrt{5}-2\sqrt{3})^2$
(I don't have a particular method for this one... maybe solve for a and b in : a²+b²=17 and ab=2sqrt(15))

Hence :
$\sqrt{-\sqrt{3}+\sqrt{4+\sqrt{5}+\sqrt{17-4\sqrt{15}}}}=\sqrt{-\sqrt{3}+\sqrt{4+\sqrt{5}+|\sqrt{5}-2\sqrt{3}|}}$

But $2 \sqrt{3}=\sqrt{12}>\sqrt{5}$
So $|\sqrt{5}-2\sqrt{3}|=2\sqrt{3}-\sqrt{5}$

This gives :
$\sqrt{-\sqrt{3}+\sqrt{4+\sqrt{5}+2\sqrt{3}-\sqrt{5}}}=\sqrt{-\sqrt{3}+\sqrt{4+2\sqrt{3}}}$

Note that $4+2\sqrt{3}=(\sqrt{3})^2+1+2 \sqrt{3}=(\sqrt{3}+1)^2$

So your expression is :
$\sqrt{-\sqrt{3}+\sqrt{3}+1}=\boxed{1}$

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