How do you simplify #sqrt(1 / 3) + sqrt( 1 / 12)#?

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Answer 1 · 2016-05-19T16:45:03

As primary root#" " sqrt(3)/2#

All possible solution #" "+-sqrt(3)/2#

When dealing with fractions it is quite often an advantage to make the denominators the same.

Multiply #1/3# by 1 but in the form of #1=4/4#

#sqrt(1/3xx4/4)+sqrt(1/12)#

#sqrt( 4xx1/12)+sqrt(1/12)#

#[sqrt(4)xxsqrt(1/12)]+sqrt(1/12)#

#2sqrt(1/12)+sqrt(1/12)#

#3sqrt(1/12)#
'~~~~~~~~~~~~~~~~~~~~~~

But it is not considered good form to have a root as the denominator.

Write as #(3sqrt(1))/sqrt(12)#

Multiply by 1 but in the form of #1=sqrt(12)/sqrt(12)#

#(3sqrt(12))/12 = sqrt(12)/4#

But 12 is #3xx4# giving

#(2sqrt(3))/4 = sqrt(3)/2#