Verified answer

√-72 = √[(36)(-1)(2)] = (√36)(√-1)(√2) = 6i√2

for a > 0, √-a = i√a ........ the i replaces the negative.

for √72, factor into √(36•2) = √36 • √2 = 6√2

put them together: √-72 = 6i√2

If you have Sqrt(-1) you can replace it with the i.

i is the solution to the problem i^2 = -1.

So if you pull out the -1 from 72 you have Sqrt(-1)*Sqrt(72) or:

i*Sqrt(72)

Then solve as a normal square root, one such way is to rewrite Sqrt(72) as Sqrt(36*2) then rewrite as 6*Sqrt(2) which makes it :

i*6*Sqrt(2).

Your second answer is wrong however because pulling everything back into the square root would give Sqrt(-1*4*6) which is not our original value. What you meant i think was i*2*Sqrt(18) which can be simplified to i*6*Sqrt(2).

first break it apart as you would any radical √(-36*2) you can bring 6 out because it is a square of 36 so 6√-2. When deal with negatives under the radical we use something called an imaginary number or i which is bascially √-1. so we can bring out i or √-1 because 1 is a perfect square. ie. 6i√2.

√-72

√(72*-1)

√72 * √-1

6√2 * i

6i√2

u break it down to its multiples and the simplest numbers are 2,2,2,3,3,...the answer wud b 6i rad 2.