How do you multiply #sqrt9*sqrt4#? Algebra Radicals and Geometry Connections Multiplication and Division of Radicals 1 Answer Don't Memorise Oct 2, 2015 #= color(blue)(6# Explanation: Method 1: #sqrt9*sqrt4= sqrt(9*4) = sqrt(36)# #sqrt36=color(blue)(6# Method 2: We know that #sqrt9=3# and, #sqrt4=2# So, #sqrt9*sqrt4= 3*2# #= color(blue)(6# 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 1091 views around the world You can reuse this answer Creative Commons License