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

1 Answer
Sep 18, 2017

Answer:

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

Explanation:

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#