How do you divide and rationalize the denominator for #2/(sqrt3+sqrt2)#? Algebra Radicals and Geometry Connections Multiplication and Division of Radicals 1 Answer Gió Apr 1, 2015 You can multiply and divide by #sqrt(3)-sqrt(2)# to get: #2/(sqrt(3)+sqrt(2))×(sqrt(3)-sqrt(2))/(sqrt(3)-sqrt(2))=(2sqrt(3)-2sqrt(2))/(3-2)=# #=2sqrt(3)-2sqrt(2)# 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 1019 views around the world You can reuse this answer Creative Commons License