$\displaystyle \frac{8}{2 - \sqrt{2}} $

is this right, $\displaystyle \frac{8}{2 - \sqrt{2}} X \frac{2 + \sqrt{2}}{2 + \sqrt{2}} $ = 16+$\displaystyle 8\sqrt{2} $, as the numerator and 6 as the dinominator?

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- Dec 18th 2009, 06:57 AMandyboy179really hard fraction question HELP
$\displaystyle \frac{8}{2 - \sqrt{2}} $

is this right, $\displaystyle \frac{8}{2 - \sqrt{2}} X \frac{2 + \sqrt{2}}{2 + \sqrt{2}} $ = 16+$\displaystyle 8\sqrt{2} $, as the numerator and 6 as the dinominator? - Dec 18th 2009, 07:04 AMdedust
- Dec 18th 2009, 07:05 AMandyboy179
just did it but i can't work out how to write it so it comes up how i want it

- Dec 18th 2009, 07:16 AMandyboy179
is what i put for the answer right? 16+http://www.mathhelpforum.com/math-he...e10ec23f-1.gif, as the numerator and 6 as the denominator?

- Dec 18th 2009, 07:28 AMdedust
- Dec 18th 2009, 07:29 AMandyboy179
- Dec 18th 2009, 07:40 AMdedust
$\displaystyle \frac{8}{2 - \sqrt{2}} \times \frac{2 + \sqrt{2}}{2 + \sqrt{2}} = \frac{8(2 + \sqrt{2})}{(2 - \sqrt{2})(2 + \sqrt{2})} $

$\displaystyle = \frac{(16 + 8\sqrt{2})}{(2^2 + 2\sqrt{2} - 2\sqrt{2} + \sqrt{2}^2)} = \frac{(16 + 8\sqrt{2})}{(4-2)} = \frac{(16 + 8\sqrt{2})}{2} = 8 + 4\sqrt{2}$

- Dec 18th 2009, 07:53 AMandyboy179
i get told to do this at school for the denominator,

Attachment 14513 - Dec 18th 2009, 07:59 AMdedust
- Dec 18th 2009, 08:02 AMandyboy179
thankyou so much