How do you multiply #sqrt6 * sqrt6#? Algebra Radicals and Geometry Connections Multiplication and Division of Radicals 2 Answers Brandon · Shantelle May 5, 2018 #6# Explanation: we know: #sqrta xx sqrtb=sqrt(ab)# #therefore# #sqrt6 xx sqrt6=sqrt36# simplifying (if possible): #sqrt36 =6# Answer link Nam D. May 6, 2018 Explanation: Given: #sqrt(6)xxsqrt6# #=(sqrt6)^2# #=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 1718 views around the world You can reuse this answer Creative Commons License