How do you simplify #sqrt(14)/sqrt(21)#?

1 Answer
Oct 1, 2015

Answer:

Note that #14=2*7and 21=3*7#

Explanation:

We can now rewrite:

#sqrt14=sqrt(2*7)=sqrt2*sqrt7#, and:
#sqrt21=sqrt(3*7)=sqrt3*sqrt7#

Putting this together, we get:

#=(sqrt2*cancelsqrt7)/(sqrt3*cancelsqrt7)=sqrt2/sqrt3#

If you multiply both halves of the fraction by #sqrt3#:

#=(sqrt3*sqrt2)/(sqrt3*sqrt3)=sqrt6/3=1/3 sqrt6#