What is the conjugate of the square root of 2 + the square root of 3 + the square root of 5? Algebra Radicals and Geometry Connections Multiplication and Division of Radicals 1 Answer George C. May 31, 2015 #sqrt(2)+sqrt(3)+sqrt(5)# does not have one conjugate. If you are trying to eliminate it from a denominator, then you need to multiply by something like: #(sqrt(2)+sqrt(3)-sqrt(5))(sqrt(2)-sqrt(3)+sqrt(5))(sqrt(2)-sqrt(3)-sqrt(5))# The product of #(sqrt(2)+sqrt(3)+sqrt(5))# and this is #-24# 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 7952 views around the world You can reuse this answer Creative Commons License