How do you simplify #( 8 + sqrt 6) /sqrt 2#? Algebra Radicals and Geometry Connections Multiplication and Division of Radicals 1 Answer Ratnesh Bhosale Mar 26, 2016 #4sqrt2+sqrt3# Explanation: #(8+sqrt6)/sqrt2# #(8+sqrt6)/sqrt2 xx sqrt2/sqrt2# #[sqrt2(8+sqrt6)]/2# #(8sqrt2+sqrt12)/2# #(8sqrt2+sqrt(4xx3))/2# #(8sqrt2+2sqrt3)/2# #[2(4sqrt2+sqrt3)]/2# #4sqrt2+sqrt3# 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 5305 views around the world You can reuse this answer Creative Commons License