How do you simplify #sqrt45divsqrt5#? Algebra Radicals and Geometry Connections Multiplication and Division of Radicals 1 Answer Don't Memorise Sep 14, 2015 #=color(blue)(3# Explanation: Prime factorising #45# to simplify #sqrt45# #sqrt45 = sqrt(3^2*5) = color(blue)(3sqrt5# The expression can now be written as: #sqrt45/sqrt5 = color(blue)((3cancelsqrt5)/cancelsqrt5# #=color(blue)(3# 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 1197 views around the world You can reuse this answer Creative Commons License