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.

Related questions
See all questions in Multiplication and Division of Radicals
Impact of this question
1984 views around the world
You can reuse this answer
Creative Commons License