How do you simplify #sqrt5 div sqrt3#?

1 Answer
Jun 28, 2017

Answer:

#sqrt(15)/3#

Explanation:

#sqrt(5)-:sqrt(3)# can be rewritten as #sqrt(5)/sqrt(3)#.

As there is a surd (radical) on the botoom, we have to rationalise the denominator, to do this we use the equation: #a/sqrt(b)-=(a*sqrt(b))/(sqrt(b)^2)=(asqrt(b))/b#.

In this case, #a=sqrt(5)# and #b=sqrt(3)#. By putting our values in we get: #sqrt(5)/sqrt(3)-=(sqrt(5)*sqrt(3))/(sqrt(3)^2)=(sqrt(5*3))/3=(sqrt(15))/3#. As none of the factors of 15 are perfect squares, this fraction cannot be simplified.