How do you simplify #sqrt(14x) div sqrt(18x)#?

1 Answer
Mar 28, 2018

Answer:

#sqrt7/3#

Explanation:

The key realization here is that we can separate the constants from the #x#s. Doing this, we have:

#(sqrt(14x))/(sqrt(18x))=(sqrt14/sqrt18)*(cancel(sqrtx/sqrtx))#

#=>sqrt14/sqrt18#

We can break up the radicals into the following:

#sqrt14/sqrt18=(sqrt(2*7))/sqrt(2*9)= (cancel(sqrt2)*sqrt7)/(cancel(sqrt2)*sqrt9)#

#sqrt2# cancels with itself, and we get

#=sqrt7/sqrt9#

Which can be simplified to:

#sqrt7/3#

Hope this helps!