How do you simplify #2/(4-sqrt6)#? Algebra Radicals and Geometry Connections Multiplication and Division of Radicals 1 Answer Konstantinos Michailidis Apr 24, 2016 It is #2/(4-sqrt6)=[2*(4+sqrt6)]/[(4-sqrt6)*(4+sqrt6)]= [2*(4+sqrt6)]/(4^2-(sqrt6)^2)= 2*(4+sqrt6)/[16-6]=2/10*(4+sqrt6)=1/5*(4+sqrt6)# 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