How do you rationalize #(2sqrt3-sqrt2)/ (5sqrt2+sqrt3)#? Algebra Radicals and Geometry Connections Multiplication and Division of Radicals 1 Answer George C. May 10, 2015 Given #(2sqrt(3)-sqrt(2))/(5sqrt(2)+sqrt(3))#, try multiplying both the top and the bottom by #(5sqrt(2)-sqrt(3))# The numerator: #(2sqrt(3)-sqrt(2))(5sqrt(2)-sqrt(3))# #=-2sqrt(3)sqrt(3)+9sqrt(2)sqrt(3)-5sqrt(2)sqrt(2)# #=-2*3+9sqrt(2*3)-5*2# #=9sqrt(6)-16# The denominator: #(5sqrt(2)+sqrt(3))(5sqrt(2)-sqrt(3))# #=25sqrt(2)sqrt(2)-sqrt(3)sqrt(3)# #=50-3# #=47#. So #(2sqrt(3)-sqrt(2))/(5sqrt(2)+sqrt(3)) = (9sqrt(6)-16)/47#. Answer link Related questions How do you simplify #\frac{2}{\sqrt{3}}#? How do you multiply and divide radicals? How do you rationalize the denominator? What is Multiplication and Division of Radicals? How do you simplify #7/(""^3sqrt(5)#? How do you multiply #(sqrt(a) +sqrt(b))(sqrt(a)-sqrt(b))#? How do you rationalize the denominator for #\frac{2x}{\sqrt{5}x}#? Do you always have to rationalize the denominator? How do you simplify #sqrt(5)sqrt(15)#? How do you simplify #(7sqrt(13) + 2sqrt(6))(2sqrt(3)+3sqrt(6))#? See all questions in Multiplication and Division of Radicals Impact of this question 1404 views around the world You can reuse this answer Creative Commons License