How do you rationalize the denominator and simplify #1/(sqrt5-3)#? Algebra Radicals and Geometry Connections Multiplication and Division of Radicals 1 Answer Hubert May 3, 2016 #1/(sqrt(5)-3)=-(sqrt(5)+3)/4# Explanation: For expressions #text(something)/(sqrt(a)-b)# we multiply the numerator and denominator by #sqrt(a)+b# so it matches the LHS of the formula #(x+y)(x-y)=x^2-y^2#. #1/(sqrt(5)-3)=1/(sqrt(5)-3) * (sqrt(5)+3)/(sqrt(5)+3)=(sqrt(5)+3)/(5-9)=-(sqrt(5)+3)/4# 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 1452 views around the world You can reuse this answer Creative Commons License