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#

Related questions
See all questions in Addition and Subtraction of Radicals
Impact of this question
2603 views around the world
You can reuse this answer
Creative Commons License