# Negative root

• Jul 30th 2012, 11:40 AM
mariusg
Negative 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 AM
Plato
Re: Negative root
Quote:

Originally Posted by mariusg
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...

Enter $-[(\sqrt{3})^2]$ and see what happens.
• Jul 30th 2012, 11:53 AM
mariusg
Re: 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 PM
mariusg
Re: 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 PM
Plato
Re: Negative root
Quote:

Originally Posted by mariusg
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?

I don't understand what you mean by that.
$-[(\sqrt{3})^2]=-3$ and $[-\sqrt{3}]^2=3$.
• Jul 30th 2012, 12:17 PM
mariusg
Re: 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 $(-\sqrt{3})^2=3$ and $(\sqrt{3})^2=3$

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

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 PM
ebaines
Re: Negative root
Quote:

Originally Posted by mariusg
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...

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 PM
mariusg
Re: 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 PM
Plato
Re: Negative root
Quote:

Originally Posted by mariusg
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.

Am i correct in assuming that in reality $(-\sqrt{3})^2=+3 or -3$? ABSOLUTELY NOT

$(-\sqrt{3})^2=+3$ period.

Here is how to do that problem: $|-\sqrt{3}+i|=\sqrt{(-\sqrt{3})^2+(1)^2}=\sqrt{3+1}=2$
• Jul 30th 2012, 12:43 PM
mariusg
Re: Negative root
Quote:

Originally Posted by Plato
$(-\sqrt{3})^2=+3$ period.

Here is how to do that problem: $|-\sqrt{3}+i|=\sqrt{(-\sqrt{3})^2+(1)^2}=\sqrt{3+1}=2$

Thanks Plato for keeping up with me,

I got a real slap in the face moment when i saw ebaines's answer. I have a tendency to overthink even the simplest things sometimes.
• Jul 31st 2012, 04:51 AM
kraj8995
Re: Negative root
This is a negative number and its square is always positive.
Like when you write
-root 3 * -root 3 =+3. always.