How do you simplify #sqrt(27/48)#? Algebra Radicals and Geometry Connections Multiplication and Division of Radicals 1 Answer Konstantinos Michailidis Sep 12, 2015 It is #3/4# Explanation: It is #sqrt(27/48)=sqrt(3^3/(3*2^4))=sqrt(3^2/2^4)=sqrt((3/2^2)^2)=3/2^2=3/4# 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 2678 views around the world You can reuse this answer Creative Commons License