How do you simplify #-20/(sqrt6 - sqrt2)#?

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Answer 1 · 2016-02-24T03:19:55

Rationalize the denominator to find that

#-20/(sqrt(6)-sqrt(2))=-5(sqrt(6)+sqrt(2))#

We will use the fact that #(a-b)(a+b) = a^2-b^2# to rationalize the denominator.

#-20/(sqrt(6)-sqrt(2)) = -(20(sqrt(6)+sqrt(2)))/((sqrt(6)-sqrt(2))(sqrt(6)+sqrt(2))#

#=-(20(sqrt(6)+sqrt(2)))/((sqrt(6))^2-(sqrt(2))^2)#

#=-(20(sqrt(6)+sqrt(2)))/(6-2)#

#=-(20(sqrt(6)+sqrt(2)))/4#

#=-5(sqrt(6)+sqrt(2))#