How do you rationalize the denominator and simplify (5sqrt3-3sqrt2)/(3sqrt2-2sqrt3)5√3−3√23√2−2√3?

2 Answers
Konstantinos Michailidis
Mar 18, 2016

Rewrite this as follows

(5sqrt3-3sqrt2)/(3sqrt2-2sqrt3)=[sqrt3*(5-sqrt3*sqrt2)]/[sqrt3*sqrt2*[sqrt3-sqrt2]]= 1/sqrt2*[5-sqrt3*sqrt2]/[sqrt3-sqrt2]5√3−3√23√2−2√3=√3⋅(5−√3⋅√2)√3⋅√2⋅[√3−√2]=1√2⋅5−√3⋅√2√3−√2

Multiply denominator and nominator with sqrt3+sqrt2√3+√2 hence

1/sqrt2*[5-sqrt3*sqrt2]/[sqrt3-sqrt2]=1/sqrt2*[(5-sqrt3*sqrt2)*(sqrt3+sqrt2)]/[(sqrt3-sqrt2)*(sqrt3+sqrt2)]= 1/(sqrt2)*[(5-sqrt3*sqrt2)*(sqrt3+sqrt2)]= 1/(sqrt2)*[2sqrt2+3sqrt3]1√2⋅5−√3⋅√2√3−√2=1√2⋅(5−√3⋅√2)⋅(√3+√2)(√3−√2)⋅(√3+√2)=1√2⋅[(5−√3⋅√2)⋅(√3+√2)]=1√2⋅[2√2+3√3]

David Britz
Mar 18, 2016

Multiply the numerator & denominator by the conjugate of the denominator.

Explanation:

Anytime you have "square root plus something" in the denominator, you multiply both num & denom by it's conjugate, i.e. change the sign in the middle.

3sqrt(2)+2sqrt(3)3√2+2√3

So the new denominator is
( 3sqrt(2)-2sqrt(3)3√2−2√3 ) * (3sqrt(2)+2sqrt(3)3√2+2√3) = 9(2)-4(3)=69(2)−4(3)=6, using (A+B)(A-B) = A^2-B^2(A+B)(A−B)=A2−B2

The new numerator is (5sqrt(3)-3sqrt(2))(3sqrt(2)+2sqrt(3))(5√3−3√2)(3√2+2√3)
which (after FOIL) is
15sqrt(6)+10(3)-9(2)-6sqrt(6)15√6+10(3)−9(2)−6√6 = =9sqrt(6)+12=9√6+12
So over the new denominator we have
=(9sqrt(6)+12)/6=9√6+126=(3sqrt(6)+4)/23√6+42

Related questions
See all questions in Multiplication and Division of Radicals
Impact of this question
2209 views around the world
You can reuse this answer
Creative Commons License