# Surds problem, please explain how?

• Jun 3rd 2008, 02:39 AM
joz31
Hi, could anyone please explain to me how to do this?

Simplify as far as possible.

2√2 ......3-2√2
----- - ---------
√6 ...... √2

• Jun 3rd 2008, 03:06 AM
Soroban
Hello, joz31!

Quote:

Simplify as far as possible: .$\displaystyle \frac{2\sqrt{2}}{\sqrt{6}} - \frac{3-2\sqrt{2}}{\sqrt{2}}$

Reduce the first fraction: . $\displaystyle \frac{2}{\sqrt{3}} - \frac{3-2\sqrt{2}}{\sqrt{2}}$

Rationalize: . $\displaystyle {\color{blue}\frac{\sqrt{3}}{\sqrt{3}}}\cdot\frac{ 2}{\sqrt{3}} \:- \:{\color{blue}\frac{\sqrt{2}}{\sqrt{2}}}\cdot\fra c{3-2\sqrt{2}}{\sqrt{2}} \;\;=\;\;\frac{2\sqrt{3}}{3} \,- \,\frac{3\sqrt{2}-4}{2}$

Get common denominator: . $\displaystyle {\color{blue}\frac{2}{2}}\cdot\frac{2\sqrt{3}}{3} \:- \:{\color{blue}\frac{3}{3}}\cdot\frac{3\sqrt{2} - 4}{2} \;\;=\;\;\frac{4\sqrt{3}}{6} - \frac{9\sqrt{2} - 12}{6}$

Therefore: . $\displaystyle \frac{4\sqrt{3} - 9\sqrt{2} + 12}{6}$

• Jun 3rd 2008, 04:36 AM
joz31
Thanks heaps for the quick reply!