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How do you simplify #sqrt(5/3)#?

Answer 1

Eva Abramson

#sqrt(5/3) = sqrt(15)/3#

Note that if #b > 0# then:

#sqrt(a/b) = sqrt(a)/sqrt(b)#

The same is not true if #a > 0# and #b < 0#.

Given to simplify:

#sqrt(5/3)#

The way I have seen most people address this kind of problem is to separate the square root then rationalise the denominator by multiplying both numerator and denominator by #sqrt(3)#, so:

#sqrt(5/3) = sqrt(5)/sqrt(3) = (sqrt(5)sqrt(3))/(sqrt(3)sqrt(3)) = sqrt(15)/3#

Personally, I prefer to multiply the numerator and denominator by #3# first, in order to make the denominator into a perfect square, thus:

#sqrt(5/3) = sqrt(15/3^2) = sqrt(15)/sqrt(3^2) = sqrt(15)/3#

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Answer 2

Jace Anderson

To simplify sqrt(5/3), you can multiply the numerator and denominator by sqrt(3) to rationalize the denominator. This gives you sqrt(5/3) * sqrt(3)/sqrt(3), which simplifies to sqrt(15)/sqrt(9). Since sqrt(9) is equal to 3, the final simplified form is sqrt(15)/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.