# Thread: really hard fraction question HELP

1. ## really 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?

2. Originally Posted by andyboy179
$\displaystyle \frac{8}{2 - \sqrt{2}}$

is this right, $\displaystyle \frac{8}{2 - \sqrt{2}} \times \frac{2 + \sqrt{2}}{2 + \sqrt{2}}$ = $\displaystyle \frac{16}+ 3\sqrt{2}}$
$\displaystyle = \frac {16 + 8\sqrt{2}}{2^2 - (\sqrt{2})^2} = . . .$

multiply it term by term and then simplify

3. just did it but i can't work out how to write it so it comes up how i want it

4. is what i put for the answer right? 16+, as the numerator and 6 as the denominator?

5. Originally Posted by andyboy179
is what i put for the answer right? 16+, as the numerator and 6 as the denominator?
the denominator is $\displaystyle 2^2 - (\sqrt{2})^2 = . . .$

6. Originally Posted by dedust
the denominator is $\displaystyle 2^2 - (\sqrt{2})^2 = . . .$

how? could you please explain it?

7. Originally Posted by andyboy179
how? could you please explain it?
$\displaystyle \frac{8}{2 - \sqrt{2}} \times \frac{2 + \sqrt{2}}{2 + \sqrt{2}} = \frac{8(2 + \sqrt{2})}{(2 - \sqrt{2})(2 + \sqrt{2})}$
$\displaystyle = \frac{(16 + 8\sqrt{2})}{(2^2 + 2\sqrt{2} - 2\sqrt{2} + \sqrt{2}^2)} = \frac{(16 + 8\sqrt{2})}{(4-2)} = \frac{(16 + 8\sqrt{2})}{2} = 8 + 4\sqrt{2}$

8. i get told to do this at school for the denominator,

9. Originally Posted by andyboy179
i get told to do this at school for the denominator,

the last entry in the 2nd column should be $\displaystyle -2$, because
$\displaystyle (-\sqrt{2}) \times (+\sqrt{2}) = -{2}$

10. thankyou so much