How do you do this problem? The square root of (2) + the square root of (2/9)

I know the answer is 4 square root of 2 over 3 but i don't know how they got that answer...

√2+ √(2/9)=
√2+√2 / 3 =
3√2 / 3 + √2 / 3 =
(3√2 + √2) / 3=

4√2 / 3

You get a common denominator after making the √9 in the denominator a normal 3. Then it's just fraction addition.

Square root is the same as sqrt()

The first thing you do is look for perfect squares in both terms...see any?

The square root of 9 is 3 so you have [sqrt(2)]+[sqrt(2)]/3. Next you need to add these together...do you see any way that might be made easier?

Multiplying by sqrt(2)/sqrt(2) gives you sqrt(2)*(1+1/3). Can you now add those fractions?

1+1/3 is 4/3 so you get sqrt(2)*4/3 which is your answer.

The square root of (2/9) is the same as the square root of 2, divided by 3. So you could write it as sqrt(2) * [1 + 1/3] = sqrt(2) * ( 4/3 ), or 4sqrt(2 / 9) if you put the 3 back in the square root sign.