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

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Answer 1 · 2018-03-28T00:35:23

#sqrt7/3#

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!