[SOLVED] Basic math taken to a next level...

**FifthRider** · June 18th 2009, 12:01 AM

Hello everyone!

During my revision for my June exams (first year engineering) I came across this problem:

$ \frac{2\sqrt{2} + 2 \sqrt{6}}{\sqrt{2+\sqrt{3}}} $

Simple math as it seems it kept me busy for 5 hours and stil no results. I know the answer is 4, but the catch is we are not allowed to use calculator. So I'm not interested in an answer, I'm interested in a method.

Any suggestions please?

I've already found multiple ways of simplifying it, so that there is no division by a square root but the method is still eluding me...

**Stroodle** · June 18th 2009, 12:15 AM

Is that $\frac{2\sqrt 2 +2\sqrt 6}{\sqrt2 +\sqrt3}$ ?
Cause that doesn't equal $\frac{4}{3}$

**flyingsquirrel** · June 18th 2009, 12:16 AM

Hi,

Originally Posted by FifthRider

During my revision for my June exams (first year engineering) I came across this problem:

$ [2*sqrt(2) + 2*sqrt(6)]/[sqrt(2 + sqrt(3))] $

Simple math as it seems it kept me busy for 5 hours and stil no results. I know the answer is $4/3$ , but the catch is we are not allowed to use calculator. So I'm not interested in an answer, I'm interested in a method.

Let $A=\frac{2\sqrt{2}+2\sqrt{6}}{\sqrt{2+\sqrt{3}}}$ . We have $A=2\sqrt{2}\cdot \frac{1+\sqrt{3}}{\sqrt{2+\sqrt{3}}}$ hence $A^2=8\cdot \frac{1+2\sqrt{3}+3}{2+\sqrt{3}}=8\times 2 =16$ which gives us $A=+\sqrt{A^2}=4$ .

**FifthRider** · June 18th 2009, 12:29 AM

Thanks, feel like kicking myself in the butt. Did I mention that the 5 hours I spent on it was this morning from 12-5? Might explain the elusive answer...

Thanks again!

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