How do you rationalize #(3sqrt3-2sqrt2)/ (3sqrt3+2sqrt2)#?

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Answer 1 · 2015-05-17T12:21:27

Multiply both the numerator (top) and denominator (bottom) by the conjugate of the denominator:

#(3sqrt(3)-2sqrt(2))/(3sqrt(3)+2sqrt(2))#

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

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

#=(27-12sqrt(6)+8)/(27-8)#

#=(35-12sqrt(6))/19#

The reason you can eliminate the square roots from the denominator in this way is because #(a+b)(a-b)=a^2-b^2# for any #a# and #b#.