How do you simplify #sqrt(6x) divsqrt24#? Algebra Radicals and Geometry Connections Multiplication and Division of Radicals 1 Answer Arunraju Naspuri Sep 22, 2015 Simplification is done by #sqrt(a)/sqrt(b)=sqrt(a/b)# and division. Explanation: We know that #sqrt(a)/sqrt(b)=sqrt(a/b)# Let #a=6x# and #b=24# #=>sqrt(6x)/sqrt(24)=sqrt((6x)/24)# #=>sqrt(6x)/sqrt(24)=sqrt((x)/4)# #=>sqrt(6x)/sqrt(24)=((x)/4)^(1/2)# 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 1336 views around the world You can reuse this answer Creative Commons License