How do you simplify #sqrt2/sqrt6#?

1 Answer
Feb 7, 2017

Answer:

Rationalize the denominator to get #sqrt3/3#

Explanation:

Start by multiplying the fraction by #sqrt6/sqrt6#,

#(sqrt6xxsqrt2)/(sqrt6xxsqrt6" "# Simplify to get

#sqrt12/6#

#=(sqrt4 xx sqrt3)/6" "# Finally, work out #sqrt4# in the numerator:

#=(2sqrt3)/6 = sqrt3/3#

Remember that #sqrtx/sqrty=sqrt(xy)/y#