How do you simplify #sqrt(6)/(4+sqrt(2))#?

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Answer 1 · 2016-03-03T10:46:00

#(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:

#(4-sqrt(2))/(4-sqrt(2))#

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

Now:

#sqrt6/(4+sqrt(2))*(4-sqrt(2))/(4-sqrt(2))#

#=(4sqrt(6)-sqrt2*sqrt6)/(16+4sqrt(2)-4sqrt(2)-2#

#=(4sqrt(6)-sqrt12)/14#

Transform the #sqrt(12)# term into a surd like so:

#=(4sqrt(6)-2sqrt(3))/14=(2sqrt(6)-sqrt(3))/7#