How do you simplify #(sqrt 3 -sqrt 6) / (sqrt 3 +sqrt6)#?

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Answer 1 · 2016-02-10T20:19:47

#=-3+2sqrt(2)#

When you have a sum of two square roots, the trick is to multiply by the equivalent subtraction:

#(sqrt(3)-sqrt(6))/(sqrt(3)+sqrt(6))#

#=(sqrt(3)-sqrt(6))/(sqrt(3)+sqrt(6))*(sqrt(3)-sqrt(6))/(sqrt(3)-sqrt(6))=#

#=((sqrt(3))^2-2*sqrt(3)*sqrt(6)+(sqrt(6))^2)/((sqrt(3))^2-(sqrt(6))^2#

#=(3-2sqrt(18)+6)/(3-6)#

#=(9-2*sqrt(9*2))/-3#

#=(9-2*3sqrt(2))/-3#

#=-3+2sqrt(2)#