what is square root of (-8)^2 ?

WolframAlpha, Matlab, Mathcad and Maple all give the following:

.

\({({(-8)^2})^{1\over2}}={(-8)^{1}}=-8\)

I would say that since \({(-8)^2}={8^2}\)

8 is also a solution.

what is square root of (-8)^2 ?

\((-8)^2=64\\ \sqrt{64}=+8\)

+8 IS THE ONLY SOLUTION

You do make a very interesting point answering guest.

\(\sqrt{(-8)^2}=\sqrt{64}=+8\\ BUT\\ \mbox{If you use index laws then}\\ \sqrt{(-8)^2}= ((-8)^2)^{0.5} =(-8)^{2*0.5}=(-8)^1=-8 \)

I still think the answer is +8 but I would like other mathematicians to comment on this.    

It is interesting when these types of contradictions spring up :)

Thanks Alan and Michael  :)

I see I sparked some discussion. So the difference is in paying close attention as to exactly how you formulate the problem.

Thanks all.

Here's another way to consider this :

l a l   =  √(a^2)       

If we let a = -8, we have

l a l   = l -8 l  =   √ [ (-8)^2 ]   = √ [64]   =   8