How do you simplify #(sqrt5+3)/(4-sqrt5)#? Algebra Radicals and Geometry Connections Multiplication and Division of Radicals 1 Answer Nityananda Sep 20, 2016 #17/11 + [7sqrt5]/11# Explanation: #[sqrt5 + 3]/[4 - sqrt5] = [sqrt5 + 3]/[4 - sqrt5] * [4 + sqrt5]/[4 + sqrt5] = [4sqrt5 + 12 + 5 + 3sqrt5]/[16 - 5]=[17 + 7sqrt5]/11 = 17/11 +[ 7sqrt5]/11# 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 1005 views around the world You can reuse this answer Creative Commons License