Simplify me

Since the root is the most outer calculation, the power is executed first, that's why the result equals:
√(16)
If it were reversed:
√(-4)²
You'd have a little problem on representing it with a real number.

EDIT:
A more valid reasoning would be to state that the power operator is not defined to be used with a negative and a non-natural (+0) operand placed in that order. Don't currently have possibilities of cooking some impressive math expression from that, sorry.

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Wolfram disagrees: http://www.wolframalpha.com/input/?i=sqrt%28%28-4%29%29^2

Noticed it at the moment I posted it, too. Edited a bit.

EDIT:
Wolfram is pretty damn impressive.

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the power operator is not defined to be used with a negative and a non-natural (+0) operand placed in that order.

Uhh... Exponentiation is defined for both zero exponents and bases.

You can exponentiate negatives. The way I do it is to ignore the minus sign, do the exponentiation as normal, and then add a minus sign if the exponent is an odd number:
-2¹ = -2
-2² = 4
-2³ = -8
-2⁴ = 16

Kyon wrote:
Wolfram is pretty damn impressive.

Wolfram|Alpha is excellent, it can do so much. Try "bacon as mass of solar system" ( http://www.wolframalpha.com/input/?i=bacon+as+mass+of+solar+system ).

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That's a lot of bacon.
Yes, WA is cool, but not as cool as running it on your own computer.

Can you do such a thing?

Sure. CAS. Today I fed Maxima (A-(aB+C))*((D-(aB+C)))=0 (upper case are R^3 and lower case are R) and it gave me the right values for a. CASs never cease to amaze me.

tition (887)

Sure. CAS. Today I fed Maxima (A-(aB+C))*((D-(aB+C)))=0 (upper case are R^3 and lower case are R) and it gave me the right values for a. CASs never cease to amaze me.

I don't get it, what exactly did you do?

Anyways, this deal with the square root: both answers - sqrt(16)=+4 and -4 - are correct.

It depends on your definition of the sqrt() function.

I don't get it, what exactly did you do?

I was trying to find the two points P on the line aB+C such that the triangle APD was a right triangle.

Anyways, this deal with the square root: both answers - sqrt(16)=+4 and -4 - are correct.

It depends on your definition of the sqrt() function.

But √((x)²) = |x|

jsmith (5804)

No, sqrt(z) for all z > 0 has both a positive and negative solution.

z^2 - 4 = 0
z^2 = 4
sqrt(z^2) = sqrt(4)
z = +/- 2

Which is obvious since the hyperbola must intersect the axis in two places

That's just a rephrasing of what I said.
z = +/- 2 <=> |z|=2

closed account (S6k9GNh0)

The sqrt of -4 to the power of 2 is 4 isn't it? Or is that just the dumb obvious mistake?

-4 to power of 2 equal 16.
sqrt of 16 is 4.

And isn't it the sqrt of negative numbers that have multiple solutions?

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@computerquip,
You're right. This is how I did it:

I wrote:
√((-4)²) = ((-4)²)^1/2 = 16^1/2 = +4

Bazzy (6281)

The square root of negative numbers have an imaginary solution

That's because:
√(-4) = √(i²4)
Right?

Bazzy (6281)

It's because (4i)² = -16