How do you simplify $\frac{\sqrt{6}}{4 + \sqrt{2}}$ $sqrt(6)/(4+sqrt(2))$ ?

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Answer 1 · 2016-03-03T10:46:00

$\frac{2 \sqrt{6} - \sqrt{3}}{6}$ $(2sqrt(6)-sqrt(3))/6$

The idea is to rationalise the denominator, we can do this by multiplying the top and the bottom of the fraction by the conjugate of the bottom:

$\frac{4 - \sqrt{2}}{4 - \sqrt{2}}$ $(4-sqrt(2))/(4-sqrt(2))$

We have just took the negative of the square root term.

Now:

$\frac{\sqrt{6}}{4 + \sqrt{2}} \cdot \frac{4 - \sqrt{2}}{4 - \sqrt{2}}$ $sqrt6/(4+sqrt(2))*(4-sqrt(2))/(4-sqrt(2))$

$= \frac{4 \sqrt{6} - \sqrt{2} \cdot \sqrt{6}}{16 + 4 \sqrt{2} - 4 \sqrt{2} - 2}$ $=(4sqrt(6)-sqrt2*sqrt6)/(16+4sqrt(2)-4sqrt(2)-2$

$= \frac{4 \sqrt{6} - \sqrt{12}}{14}$ $=(4sqrt(6)-sqrt12)/14$

Transform the $\sqrt{12}$ $sqrt(12)$ term into a surd like so:

$= \frac{4 \sqrt{6} - 2 \sqrt{3}}{14} = \frac{2 \sqrt{6} - \sqrt{3}}{7}$ $=(4sqrt(6)-2sqrt(3))/14=(2sqrt(6)-sqrt(3))/7$

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