How to add and subtract radical equations #(3sqrt2)/9 +1/(2sqrt32)+sqrt9/8#?

1 Answer
Apr 3, 2015

We first simplify the individual terms as much as possible

#(3sqrt2)/9=sqrt2/3# (divide upper and lower by #3#)

#32=2*16=2*4^2->sqrt32=4sqrt2->#
#1/(2sqrt32)=1/(8sqrt2#

#9=3^2->sqrt9/8=3/8#

Our problem has now evolved into:

#sqrt2/3 + 1/(8sqrt2) +3/8#

We need to take care of the second term by multiplying (above and below) by #sqrt2#

#1/(8sqrt2) *sqrt2/sqrt2=sqrt2/(8*2)=sqrt2/16#

Now we need to get all numerators alike; we'll have to get them to #48# (= smallest common multiple)

#(16*sqrt2)/(16*3) + (3*sqrt2)/(3*16) + (6*3)/(6*8)=#

#(16sqrt2+3sqrt2+18)/48=(19sqrt2+18)/48=19/48sqrt2 +3/8#