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

2 Answers

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]#

Mar 18, 2016

Answer:

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#