How do you rationalize #(sqrt3-sqrt2)/ ( sqrt3+sqrt2)#? Algebra Radicals and Geometry Connections Multiplication and Division of Radicals 1 Answer GiĆ³ May 16, 2015 You can multiply and divide by: #sqrt(3)-sqrt(2)# to get: #(sqrt(3)-sqrt(2))/(sqrt(3)+sqrt(2))*(sqrt(3)-sqrt(2))/(sqrt(3)-sqrt(2))=# #=(sqrt(3)-sqrt(2))^2/(3-2)=3-2sqrt(6)+2=5-2sqrt(6)# 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 1920 views around the world You can reuse this answer Creative Commons License