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

Answer 1

#sqrt(245/3)=(7sqrt15)/3#

#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#

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Answer 2

Isaiah Anthony

To rationalize the denominator and simplify sqrt(245/3), we can multiply both the numerator and denominator by the conjugate of the denominator. The conjugate of sqrt(3) is also sqrt(3).

By multiplying sqrt(245/3) by sqrt(3)/sqrt(3), we get sqrt(245*3)/(sqrt(3)*sqrt(3)).

This simplifies to sqrt(735)/3.

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Answer from HIX Tutor

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.