How do you simplify #sqrt36/sqrt3#? Algebra Radicals and Geometry Connections Multiplication and Division of Radicals 1 Answer Don't Memorise Jan 28, 2016 #=2sqrt3# Explanation: #sqrt36/sqrt3# We know that #sqrt36= 6# So, #sqrt36/sqrt3 = color(blue)(6)/sqrt3# Now we rationalise the denominator: #(6)/sqrt3 =((6) xx sqrt3) /(sqrt3 xx sqrt3 # #=(6sqrt3)/(sqrt3 xx sqrt3# #=(6sqrt3)/ 3# #=(cancel6sqrt3)/ cancel3# #=2sqrt3# 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 1035 views around the world You can reuse this answer Creative Commons License