How do you simplify #4 /(sqrt7 + sqrt 3)#? Algebra Radicals and Geometry Connections Multiplication and Division of Radicals 1 Answer MeneerNask Apr 3, 2017 You multiply the both halves by #sqrt7-sqrt3# and use the special product #(A+B)(A-B)=A^2-B^2# Explanation: #=4/(sqrt7-sqrt3)xx(sqrt7-sqrt3)/(sqrt7-sqrt3)# #=(4(sqrt7-sqrt3))/((sqrt7+sqrt3)(sqrt7-sqrt3))# #=(4(sqrt7-sqrt3))/(sqrt7^2-sqrt3^2)=(4(sqrt7-sqrt3))/(7-3)=# #=(cancel4(sqrt7-sqrt3))/cancel4=sqrt7-sqrt3# 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 5367 views around the world You can reuse this answer Creative Commons License