How do you rationalize the denominator #2/(sqrt[3] +sqrt[2])#? Algebra Radicals and Geometry Connections Multiplication and Division of Radicals 1 Answer GiĆ³ Apr 12, 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))=# #=(2*(sqrt(3)-sqrt(2)))/(3-2)=2*(sqrt(3)-sqrt(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 2829 views around the world You can reuse this answer Creative Commons License