How do you simplify #sqrt3 div(sqrt12 - sqrt5)#? Algebra Radicals and Geometry Connections Multiplication and Division of Radicals 1 Answer Bhavin B. May 15, 2018 #(sqrt3)/(2sqrt3-sqrt5)# Explanation: First, we find the factors of #sqrt12# and simplify it further. #sqrt12# = #sqrt(4 xx 3# = #sqrt4xxsqrt3# = #2xxsqrt3# #(sqrt3)/(sqrt12 - sqrt5)# ----->we know that #sqrt12# is #2xxsqrt3# and substitute that for #sqrt12# as shown below: #(sqrt3)/(2xxsqrt3 -sqrt5)# = #(sqrt3)/(2sqrt3-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 2016 views around the world You can reuse this answer Creative Commons License