How do you simplify #2sqrt3/sqrt6##? Algebra Radicals and Geometry Connections Multiplication and Division of Radicals 1 Answer Elvan Burcu K. May 28, 2015 #2* sqrt3/sqrt6 = 2* sqrt3/sqrt(3*2) = 2* sqrtcancel3/sqrt(cancel3*2) = 2/sqrt2 =sqrt(2*2)/ sqrt2 = sqrt(cancel2*2)/sqrtcancel2= sqrt2# 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 1362 views around the world You can reuse this answer Creative Commons License