How do you simplify #sqrt(2x ) div sqrt(9x)#?

1 Answer
Sep 20, 2015

Answer:

#sqrt2/3#

Explanation:

Rewrite in fractional form:
#sqrt(2x) -: sqrt(9x) = sqrt(2x)/sqrt(9x)#

Remember the rule:
#sqrt(ab)=sqrt(a)sqrt(b)#

so breaking down the roots, we get:
#= (sqrt(2)sqrt(x))/(sqrt(9)sqrt(x))#

Now, #sqrt(x)/sqrt(x)=1# so that's gone, and #sqrt(9)=3#, so we end up having:
#= sqrt2/3#