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

• Jun 18th 2009, 12:01 AM
FifthRider
[SOLVED] Basic math taken to a next level...
Hello everyone!

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

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

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.

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...
• Jun 18th 2009, 12:15 AM
Stroodle
Is that $\displaystyle \frac{2\sqrt 2 +2\sqrt 6}{\sqrt2 +\sqrt3}$ ?
Cause that doesn't equal $\displaystyle \frac{4}{3}$
• Jun 18th 2009, 12:16 AM
flyingsquirrel
Hi,
Quote:

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

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

Simple math as it seems it kept me busy for 5 hours and stil no results. I know the answer is $\displaystyle 4/3$ , 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 $\displaystyle A=\frac{2\sqrt{2}+2\sqrt{6}}{\sqrt{2+\sqrt{3}}}$. We have $\displaystyle A=2\sqrt{2}\cdot \frac{1+\sqrt{3}}{\sqrt{2+\sqrt{3}}}$ hence $\displaystyle A^2=8\cdot \frac{1+2\sqrt{3}+3}{2+\sqrt{3}}=8\times 2 =16$ which gives us $\displaystyle A=+\sqrt{A^2}=4$.
• Jun 18th 2009, 12:29 AM
FifthRider
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!