How do you simplify #(8sqrt 2)/ (sqrt 3)#?

1 Answer
Feb 10, 2016

Answer:

#(8sqrt6)/3#

Explanation:

We want to "rationalize" the denominator--that is, get all square roots out of the denominator. We can do this by multiplying the fraction by #sqrt3"/"sqrt3#.

#=(8sqrt2)/sqrt3*sqrt3/sqrt3=(8sqrt2sqrt3)/3#

To simplify the numerator, recall that when #a,b# are positive, #sqrtasqrtb=sqrt(ab)#. Hence

#=(8sqrt(2xx3))/3=(8sqrt6)/3#