How do you evaluate #12\sqrt { 2} \div 2\sqrt { 27}#?

1 Answer
Jun 11, 2017

It's #(2sqrt6)/3#.

Explanation:

Let's start with the original expression:

#(12*sqrt2)/(2*sqrt27)#

We can't do anything about the #sqrt2# but we can split the #sqrt27# into #sqrt3# and #sqrt9#. We notice that we can simplify the #sqrt9# into #3#. Now, let's combine everything together:

#(12*sqrt2)/(2*sqrt3*sqrt9)=(12sqrt2)/(2*sqrt3*3)=(12sqrt2)/(6sqrt3)#

Now what? Well, we see that we can split the final expression into #12/6# and #sqrt2/sqrt3#. And of course we can simplify the #12/6# into #2#:

#(12sqrt2)/(6sqrt3)=12/6*sqrt2/sqrt3=2*sqrt2/sqrt3#

But how do we simplify #sqrt2/sqrt3#? There's a trick that we can use and it's called rationalizing the denominator. Essentially we multiply #sqrt2/sqrt3# by #sqrt3/sqrt3# and get this:

#sqrt2/sqrt3*sqrt3/sqrt3=(sqrt2*sqrt3)/(sqrt3*sqrt3)=sqrt6/sqrt9=sqrt6/3#

There! Now we can go back to the equation and continue working:

#2*sqrt2/sqrt3=2*sqrt6/3=(2sqrt6)/3#

Hope this helped!