How do you simplify #(3sqrt28)/sqrt48#? Algebra Radicals and Geometry Connections Multiplication and Division of Radicals 1 Answer José F. Apr 24, 2016 #sqrt(7)/2# Explanation: #(3sqrt(28))/sqrt(48)=(3sqrt(2^2*7))/sqrt(2^4*3)=(3*2sqrt(7))/(2^2sqrt(3))=3/2*sqrt(7)/sqrt(3)*sqrt(3)/sqrt(3)=3/(2*3)sqrt(7)=sqrt(7)/2# 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 914 views around the world You can reuse this answer Creative Commons License