How do you add #sqrt(2/3) + sqrt(4/3)#?

2 Answers
Joe D.
Apr 2, 2015

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

#(sqrt(2/3) + sqrt(4/3))^2 = c^2#

#2/3 + 2 * sqrt(2/3) * sqrt(4/3) + 4/3= c^2#

#6/3 + 2 * sqrt(2/3 * 4/3) = c^2#

#2 + 2 * sqrt(2 * 4/9) = c^2#

#2 + 2 * sqrt(2 * 2^2/3^2) = c^2#

#2 + 2 * 2/3 * sqrt(2) = c^2#

#2 + (4sqrt(2))/3 = c^2#

#(6 + 4sqrt(2))/3 = c^2#

#c = sqrt((6 + 4sqrt(2))/3)#

The result is #c#.

As you can see, doing this hardwork is meaningless. Root operator is prior to the addition. The result is more complex than the question. So we can easily tell that the question is already in the simplest form

Jim H
Apr 2, 2015

#sqrt (2/3) =sqrt2/sqrt3= sqrt2/sqrt3 *sqrt3/sqrt3 =sqrt6/3#

#sqrt(4/3)= sqrt4/sqrt3= (2sqrt3)/3#

So
#sqrt (2/3)+sqrt (4/3) = sqrt6/3 + (2sqrt3)/3 =(sqrt6+2sqrt3)/3#

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