How do you simplify #sqrt(7/3) *sqrt(7/3)#? Algebra Radicals and Geometry Connections Multiplication and Division of Radicals 1 Answer Don't Memorise May 3, 2015 We know that #color(blue)(sqrta*sqrtb = sqrt(ab)# Hence #sqrt(7/3) *sqrt(7/3) = sqrt((7/3)(7/3))# # = sqrt(7^2/3^2)# # = sqrt((7/3)^2)# # = color(green)(7/3)# #sqrt(7/3) *sqrt(7/3) = 7/3# 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 1156 views around the world You can reuse this answer Creative Commons License