How do you simplify #sqrt4/sqrt36#? Algebra Radicals and Geometry Connections Multiplication and Division of Radicals 1 Answer Shwetank Mauria Sep 25, 2016 #sqrt4/sqrt36=1/3# Explanation: #sqrt4/sqrt36# = #sqrt(2xx2)/sqrt(2xx2xx3xx3)# = #sqrt(ul(2xx2))/sqrt(ul(2xx2)xxul(3xx3))# = #2/(2xx3)# = #(1cancel2)/(cancel2xx3)# = #1/3# 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 854 views around the world You can reuse this answer Creative Commons License