How do you simplify #sqrt(54y^2)/ sqrt(6y)#? Algebra Radicals and Geometry Connections Multiplication and Division of Radicals 1 Answer Nallasivam V Oct 1, 2015 #3.sqrt y# Explanation: #sqrt(54y^2)/sqrt(6y)# #(sqrt(6).sqrt(9).sqrt(y^2))/(sqrt(6) sqrt(y))# #(cancel(sqrt(6)).sqrt(9).sqrt(y^2))/(cancelsqrt(6) sqrt(y))# #(.sqrt(9).sqrt(y^2))/( sqrt(y))# #(3.y)/( sqrt(y))# #3.sqrt y# 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 1216 views around the world You can reuse this answer Creative Commons License