Author Topic: help (Read 1001 times) Tweet Share

vcestudent94 · « **on:** November 07, 2012, 09:25:27 pm »

How do you convert: sqrt(sqrt(3)+2)/2 to--> (sqrt(6)+sqrt(2))/4
I hope thats clear

BubbleWrapMan · « **Reply #1 on:** November 07, 2012, 09:45:41 pm »

$\frac{\sqrt{\sqrt{3}+2}}{2}=\frac{2\sqrt{\sqrt{3}+2}}{4}=\frac{\sqrt{4(\sqrt{3}+2)}}{4}=\frac{\sqrt{4\sqrt{3}+8}}{4}=\frac{\sqrt{6+4\sqrt{3}+2}}{4}=\frac{\sqrt{(\sqrt{6})^{2}+2\cdot\sqrt{2}\cdot\sqrt{2}\cdot\sqrt{3}+(\sqrt{2})^{2}}}{4}=\frac{\sqrt{(\sqrt{6})^{2}+2\cdot\sqrt{6}\cdot\sqrt{2}+(\sqrt{2})^{2}}}{4}=\frac{\sqrt{(\sqrt{6}+\sqrt{2})^{2}}}{4}=\frac{\sqrt{6}+\sqrt{2}}{4}$

But needless to say you wouldn't have to do that in tech-free

HighLatency · « **Reply #2 on:** November 07, 2012, 09:53:56 pm »

Convert sqrt(3)+2 into polar form
Sqrt the polar form using the theorem thingo
Convert the polar form back into Cartesian
Put into original and bam, answer

BubbleWrapMan · « **Reply #3 on:** November 07, 2012, 10:03:22 pm »

Wait, what's the polar form of that

HighLatency · « **Reply #4 on:** November 07, 2012, 11:55:01 pm »

Quote from: ClimbTooHigh on November 07, 2012, 10:03:22 pm

Wait, what's the polar form of that

O crap...
Yeah that doesn't work xD

Biceps · « **Reply #5 on:** November 08, 2012, 09:06:52 am »

LOL i was waiting for that :PP

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