How do you rationalize (3sqrt3-2sqrt2)/ (3sqrt3+2sqrt2)3√3−2√23√3+2√2?

1 Answer
George C.
May 17, 2015

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

(3sqrt(3)-2sqrt(2))/(3sqrt(3)+2sqrt(2))3√3−2√23√3+2√2

= ((3sqrt(3)-2sqrt(2))(3sqrt(3)-2sqrt(2)))/((3sqrt(3)+2sqrt(2))(3sqrt(3)-2sqrt(2)))=(3√3−2√2)(3√3−2√2)(3√3+2√2)(3√3−2√2)

= ((3sqrt(3))^2-2(3sqrt(3))(2sqrt(2))+(2sqrt(2))^2)/((3sqrt(3))^2-(2sqrt(2))^2)=(3√3)2−2(3√3)(2√2)+(2√2)2(3√3)2−(2√2)2

=(27-12sqrt(6)+8)/(27-8)=27−12√6+827−8

=(35-12sqrt(6))/19=35−12√619

The reason you can eliminate the square roots from the denominator in this way is because (a+b)(a-b)=a^2-b^2(a+b)(a−b)=a2−b2 for any aa and bb.

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