How do you rationalize the denominator and simplify #(8-6sqrt5)/sqrt12#? Algebra Radicals and Geometry Connections Multiplication and Division of Radicals 1 Answer Nityananda Mar 1, 2017 #[8-6 sqrt5]/sqrt12 = [2(4-3sqrt5)]/sqrt(4xx3)# #rArr [2(4-3sqrt5)]/[2sqrt3]# #rArr [4-3sqrt5]/[sqrt3]# multiply D & N BY#sqrt3# #rArr [sqrt3(4-3sqrt5)]/[sqrt3 sqrt3]# #rArr[sqrt3(4-3sqrt5)]/3 # 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 862 views around the world You can reuse this answer Creative Commons License