Question

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

Answer 1

Rewrite this as follows

#(5sqrt3-3sqrt2)/(3sqrt2-2sqrt3)=[sqrt3*(5-sqrt3*sqrt2)]/[sqrt3*sqrt2*[sqrt3-sqrt2]]= 1/sqrt2*[5-sqrt3*sqrt2]/[sqrt3-sqrt2]#

Multiply denominator and nominator with `sqrt3+sqrt2` 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]#

Answer 2

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)`

So the new denominator is
( `3sqrt(2)-2sqrt(3)` ) * (`3sqrt(2)+2sqrt(3)`) = `9(2)-4(3)=6`, using `(A+B)(A-B) = A^2-B^2`

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