How do you simplify #(2 - sqrt5) / (3 +sqrt5)#? Algebra Radicals and Geometry Connections Multiplication and Division of Radicals 1 Answer ali ergin Apr 20, 2016 #(2-sqrt5)/(3+sqrt5)=(11-5sqrt5)/4# Explanation: #(2-sqrt5)/(3+sqrt5)# #((2-sqrt5)(3-sqrt5))/((3+sqrt5)(3-sqrt5))=(6-2sqrt5-3sqrt5+(sqrt5)^2)/(3^2-(sqrt5)^2)# #(6-5sqrt5+5)/(9-5)=(11-5sqrt5)/4# #(2-sqrt5)/(3+sqrt5)=(11-5sqrt5)/4# 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 851 views around the world You can reuse this answer Creative Commons License