Hi, could anyone please explain to me how to do this?

Simplify as far as possible.

2√2 ......3-2√2

----- - ---------

√6 ...... √2

Thanks in advance for all your help.

Printable View

- Jun 3rd 2008, 02:39 AMjoz31Surds problem, please explain how?
Hi, could anyone please explain to me how to do this?

*Simplify as far as possible*.

2√2 ......3-2√2

----- - ---------

√6 ...... √2

Thanks in advance for all your help. - Jun 3rd 2008, 03:06 AMSoroban
Hello, joz31!

Quote:

Simplify as far as possible: .$\displaystyle

\frac{2\sqrt{2}}{\sqrt{6}} - \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 AMjoz31
Thanks heaps for the quick reply!