Can anyone help me figure this problem out step by step? All help would be appreciated.

3/6-√2

√=radical

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- Apr 19th 2008, 05:14 PMtyparadiseRationalize the denominator
Can anyone help me figure this problem out step by step? All help would be appreciated.

3/6-√2

√=radical - Apr 19th 2008, 05:21 PMTheEmptySet
use the conjugate

so we have $\displaystyle 6-\sqrt{2}$ its conjugate is $\displaystyle 6+\sqrt{2}$

mult the numerator and denominator by the conjugate to get

$\displaystyle \frac{3}{6-\sqrt{2}} \cdot \frac{6+\sqrt{2}}{6+\sqrt{2}}=\frac{18+3\sqrt{2}}{ 6-2}=\frac{9}{2}+\frac{3}{4}\sqrt{2}$ - Apr 19th 2008, 05:58 PMtyparadiseRationalize the denominator
OMG Thanks.

But can you explain after you multiply by conjugate how did you get the negative for the radical. - Apr 19th 2008, 06:03 PMTheEmptySet