Ok if I enter (-sqrt(3))^2) into my calculator I get the answer 3. Where does the - go?

I know this is basic but...

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- Jul 30th 2012, 11:40 AMmariusgNegative root
Ok if I enter (-sqrt(3))^2) into my calculator I get the answer 3. Where does the - go?

I know this is basic but... - Jul 30th 2012, 11:43 AMPlatoRe: Negative root
- Jul 30th 2012, 11:53 AMmariusgRe: Negative root
I did that. And it preserved the negative sign.

But in my case I need to calculate a length of the vector and I know that one one of the short sides is -sqrt(3) (it is actually with complex numbers but I like to think of complex numbers at kind of vectors. Is that wrong?)

And I am using the c=sqrt(a^2+b^2).

Sin this case using -((sqrt(3)^2) is incorrect isn't it? - Jul 30th 2012, 12:02 PMmariusgRe: Negative root
Wait a moment, I think I understand it. Because the value of the square root can be either positive or negative. The sign becomes redundant? Can then the length/magnitude of a vector have two solutions if either a or b is a root?

- Jul 30th 2012, 12:04 PMPlatoRe: Negative root
- Jul 30th 2012, 12:17 PMmariusgRe: Negative root
Ok, i will start from scratch.

I have this problem:

Write -sqrt(3)+i in polar from. (Again translated so might be the way it is normally phrased).

What i did was trying to find the |v| value:

sqrt((-sqrt(3))^2+1^2)

Now, at the (-sqrt(3))^2 is where my question comes in.

because my calculator shows that both $\displaystyle (-\sqrt{3})^2=3$ and $\displaystyle (\sqrt{3})^2=3$

Am i correct in assuming that in reality $\displaystyle (-\sqrt{3})^2=+3 or 3$?

Are then the negative value not used because it will result in a negative magnitude and that is not allowed?

Are there cases where using the pythagorean theorem in this way will result in two possible solutions for the magnitude? - Jul 30th 2012, 12:21 PMebainesRe: Negative root
You enter sqrt(3) and the calculator says 1.732...

Then make it negative and you have -1.732..

Then square that: -1.732 x -1.732 = +3.

So yes, it's a positive result. You're basically asking why a negative times a negative is a positive. And note that if your calculator had given you the negative value of sqrt(3) you'd still get the same thing: -(-1.732) x -(-1.732) = 3. - Jul 30th 2012, 12:24 PMmariusgRe: Negative root
Right,

I should actually have tried to do that little experiment, insted of overcomplicating everything.

Thanks for youre help. - Jul 30th 2012, 12:28 PMPlatoRe: Negative root
- Jul 30th 2012, 12:43 PMmariusgRe: Negative root
- Jul 31st 2012, 04:51 AMkraj8995Re: Negative root
This is a negative number and its square is always positive.

Like when you write

-root 3 * -root 3 =+3. always.