How do you simplify #sqrt(3/4)*sqrt(4/5)#? Algebra Radicals and Geometry Connections Multiplication and Division of Radicals 1 Answer Alan P. Apr 7, 2015 Since #sqrt(a)*sqrt(b) = sqrt(a*b)# #sqrt(3/4) * sqrt(4/5)# #= sqrt( 3/cancel(4) * cancel(4)/5)# # = sqrt(3/5)# 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 1640 views around the world You can reuse this answer Creative Commons License