How do you simplify #sqrt2/sqrt6#? Algebra Radicals and Geometry Connections Multiplication and Division of Radicals 1 Answer Aaron C. · EZ as pi Feb 7, 2017 Rationalize the denominator to get #sqrt3/3# Explanation: Start by multiplying the fraction by #sqrt6/sqrt6#, #(sqrt6xxsqrt2)/(sqrt6xxsqrt6" "# Simplify to get #sqrt12/6# #=(sqrt4 xx sqrt3)/6" "# Finally, work out #sqrt4# in the numerator: #=(2sqrt3)/6 = sqrt3/3# Remember that #sqrtx/sqrty=sqrt(xy)/y# 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 1684 views around the world You can reuse this answer Creative Commons License