How do you rationalize the denominator and simplify #sqrt(245/3)#? Algebra Radicals and Geometry Connections Multiplication and Division of Radicals 1 Answer Shwetank Mauria Jul 11, 2016 #sqrt(245/3)=(7sqrt15)/3# Explanation: #sqrt(245/3)=sqrt245/sqrt3# As we have #sqrt3# in denominator, we need to multiply it by #sqrt3#, that will make the denominator #sqrt9=3# and thus rationalise the denominator. But as we multiply denominator by #sqrt3#. we should also multiply numerator by #sqrt3#. Hence, #sqrt(245/3)=sqrt245/sqrt3# = #(sqrt245×sqrt3)/(sqrt3×sqrt3# = #sqrt735/3# = #sqrt(3×5×ul(7×7))/3# = #(7sqrt15)/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 1006 views around the world You can reuse this answer Creative Commons License