How do you simplify #sqrt3/ sqrt5#? Algebra Radicals and Geometry Connections Multiplication and Division of Radicals 1 Answer Nakanojonon Jun 1, 2018 #sqrt(15)/5# Explanation: You just have to multiply both numerator #(sqrt(3))# and denominator #(sqrt(5))# with #sqrt(5)# to rationalize the form #sqrt(3)/sqrt(5)# = #sqrt(3)/sqrt(5)# * #sqrt(5)/sqrt(5)# = #sqrt(15)/sqrt(25)# = #sqrt(15)/5# 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 2786 views around the world You can reuse this answer Creative Commons License