How do you rationalize the denominator and simplify #sqrt(245/3)#?

1 Answer
Jul 11, 2016

Answer:

#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#