How do you simplify #(9 sqrt(5x^2))/( 3 sqrt(2x^4))#? Algebra Radicals and Geometry Connections Multiplication and Division of Radicals 1 Answer MeneerNask Feb 21, 2016 Remember that #sqrt(x^2)=x# and #x^4=x^2*x^2# And that #sqrt(a*b)=sqrta*sqrtb# Explanation: #=(9*sqrt5*sqrt(x^2))/(3*sqrt2*sqrt(x^2)*sqrt(x^2))# #=(cancel3*3*cancelx*sqrt5)/(cancel3*x*cancelx*sqrt2)# #=(3sqrt5)/(xsqrt2)# 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 1151 views around the world You can reuse this answer Creative Commons License