How do you simplify #sqrt63/sqrt7#? Algebra Radicals and Geometry Connections Multiplication and Division of Radicals 1 Answer sente Apr 26, 2016 #sqrt(63)/sqrt(7) = 3# Explanation: Using the property that #sqrt(a)/sqrt(b) = sqrt(a/b)# for #a>=0#, #b>0#: #sqrt(63)/sqrt(7) = sqrt(63/7) = sqrt(9) = sqrt(3^2) = 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 1066 views around the world You can reuse this answer Creative Commons License