How do you simplify #(sqrt3+2)^2#? Algebra Radicals and Geometry Connections Multiplication and Division of Radicals 1 Answer Ratnaker Mehta May 25, 2017 # 7+4sqrt3.# Explanation: Using the Expansion Formula, #(a+b)^2=a^2+2ab+b^2,# we have, #(sqrt3+2)^2=sqrt3^2+2*sqrt3*2+2^2,# # =3+4sqrt3+4,# # rArr (sqrt3+2)^2=7+4sqrt3.# 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 1124 views around the world You can reuse this answer Creative Commons License