How do you simplify #sqrt2 / (2 sqrt6-sqrt2)#? Algebra Radicals and Geometry Connections Multiplication and Division of Radicals 1 Answer F. Javier B. May 24, 2018 The goal is remove roots from denominator. See below Explanation: #sqrt2/(2sqrt6-sqrt2)=(sqrt2(2sqrt6+sqrt2))/((2sqrt6-sqrt2)(2sqrt6+sqrt2))=(2sqrt2sqrt6+2)/(24-2)=(2sqrt12+2)/22=sqrt2/11+1/11# 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 1834 views around the world You can reuse this answer Creative Commons License