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Discriminant of a cubic

  1. May 10, 2005 #1
    On mathworld's discussion of the cubic formula he has that

    "determining which roots are real and which are complex can be accomplished by noting that if the polynomial discriminant D > 0, one root is real and two are complex conjugates; if D = 0, all roots are real and at least two are equal; and if D < 0, all roots are real and unequal."

    Does that sound wrong to anyone else? It's been a while since I learned about cubic discriminants but doesn't a negative discriminant mean two complex roots?

    I had actually forgotten what a negative cubic discriminant meant so I was looking it up but this seems wrong to me. Anybody feel confident one way or the other?

    Thanks,
    Steven
     
  2. jcsd
  3. May 11, 2005 #2

    matt grime

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    The discriminant is, usually defined as:

    given a poly P, let x_1,..,x_n be its roots in some order (assume they all exist ie deg(P)=n; we're in some splitting field), then let d= prod_{i<j}(x_i-x_j}. The discriminant is D=d^2.

    However, quickly reading that page I can see where there may be confusion, for as given you're right to be confused. WHen it first mentions discriminants it links to the mathworld definition which agrees with mine, but mentions that Birkhoff and someone else use something that differs by a minus sign.

    Two scenarios:
    1 he got confused as to which convention he was using later in the piece, or

    2 they altered the definition on mathworld afterwards to agree with birkhoff's and it used to have the minus sign on the mathworld page.

    not having a copy of birkhoff i can't say which is more likely.
     
  4. May 11, 2005 #3
    Ah, right you are. I actually saw that line then dismissed it when I saw that the roots use the square root of the discriminant. Looks like the imaginary part cancels out.

    Thanks a lot,
    Steven
     
  5. May 11, 2005 #4
    matt grime:given a poly P, let x_1,..,x_n be its roots in some order (assume they all exist ie deg(P)=n; we're in some splitting field), then let d= prod_{i<j}(x_i-x_j}. The discriminant is D=d^2.

    Clearly if the roots were real, using that definition the discriminant would be positive. If there was a pair of complex, then the difference between them would be imaginary.
     
  6. May 11, 2005 #5

    matt grime

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    I'm sorry, Robert, your point was what? I thought I explained that with that definition what is written at Mathworld is wrong for that interpretation, though there appears to be other definitions that differ by a minus sign.
     
  7. May 11, 2005 #6
    My point was nothing really, I was just agreeing, not arguing with you--if that adds any substance to the subject. I too was very confused by this matter.

    However, I can now add that some writers use q^2/4 + p^3/27, where this is what is under the radical in the answer. I guess it is a horse of another color.

    A definition of that can be found:http://mathworld.wolfram.com/CubicFormula.html

    Which gives [tex] p=\frac{3a_1-a_2^2}{3}[/tex]

    [tex]q=\frac{9a_1a_2-27a_0-2a_2^3}{27}[/tex]

    Where the equation is: [tex]z^3+a_2z^2+a_1z+a_0. [/tex]

    When the value under the square root is negative, that is called the Irreducible Case! And that is the case where you have real roots! It is another use of the word discriminant. The answer there is to use DeMoivre’s Theorem..

    Cardan noticed that himself and gave the example of X^3=15X+4, where 4 is a root.
     
    Last edited: May 11, 2005
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