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Math Help - Negative root

  1. #1
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    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...
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  2. #2
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    Re: Negative root

    Quote Originally Posted by mariusg View Post
    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.
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  3. #3
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    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?
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  4. #4
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    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?
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  5. #5
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    Re: Negative root

    Quote Originally Posted by mariusg View Post
    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.
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  6. #6
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    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?
    Last edited by mariusg; July 30th 2012 at 01:17 PM.
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  7. #7
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    Re: Negative root

    Quote Originally Posted by mariusg View Post
    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.
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  8. #8
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    Re: Negative root

    Right,

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

    Thanks for youre help.
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    Re: Negative root

    Quote Originally Posted by mariusg View Post
    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
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  10. #10
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    Re: Negative root

    Quote Originally Posted by Plato View Post
    (-\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.
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  11. #11
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    Re: Negative root

    This is a negative number and its square is always positive.
    Like when you write
    -root 3 * -root 3 =+3. always.
    Last edited by skeeter; August 4th 2012 at 06:01 PM. Reason: deleted link
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