How do you simplify the expression #(sqrt2+sqrt3)^2#? Algebra Radicals and Geometry Connections Multiplication and Division of Radicals 1 Answer MeneerNask Feb 8, 2017 You work out the expression #(a+b)^2=a^2+2ab+b^2# Explanation: #=sqrt2^2+2*sqrt2*sqrt3+sqrt3^2# #=2+2*sqrt(2*3)+3# #=5+2sqrt6# 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 2288 views around the world You can reuse this answer Creative Commons License