How do you rationalize the denominator and simplify #4/(sqrt7-sqrt5)#? Algebra Radicals and Geometry Connections Multiplication and Division of Radicals 1 Answer sjc · George C. Nov 12, 2017 #=2(sqrt7+sqrt5)# Explanation: #4/(sqrt7-sqrt5)# to rationalise the denominator we make use of #(a+b)(a-b)=a^2-b^2# so #4/(sqrt7-sqrt5)= 4/(sqrt7-sqrt5)xx (sqrt7+sqrt5)/(sqrt7+sqrt5)# #=(4(sqrt7+sqrt5))/((sqrt7)^2-(sqrt5)^2)# #=(4(sqrt7+sqrt5))/(7-5)# #=(cancel(4)^2(sqrt7+sqrt5))/cancel(2)# #=2(sqrt7+sqrt5)# 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 1265 views around the world You can reuse this answer Creative Commons License