How do you rationalize #7/(sqrt3-sqrt2)#? Algebra Radicals and Geometry Connections Multiplication and Division of Radicals 1 Answer bp May 8, 2015 7#(sqrt3+sqrt2)# Multiply the numerator and the denominator by #sqrt3+sqrt2#, that would make the denominator #(sqrt3-sqrt2)(sqrt3+sqrt2)# =3-2= 1. The result would be #7(sqrt3+sqrt2)#. That is what was required. 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 1492 views around the world You can reuse this answer Creative Commons License