Is the Discriminant of a Cubic Always Negative When There are Real Roots?

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The discussion centers on the confusion surrounding the discriminant of cubic polynomials and its implications for the nature of the roots. It is clarified that a positive discriminant indicates one real root and two complex conjugates, while a zero discriminant means all roots are real with at least two equal. The key point of contention is the interpretation of a negative discriminant, which some participants argue does not align with the conventional understanding that it typically indicates complex roots. The conversation also touches on different definitions of the discriminant, suggesting that variations in terminology may contribute to the confusion. Ultimately, the discussion emphasizes the need for clarity in mathematical definitions and conventions regarding cubic discriminants.
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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
 
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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.
 
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
 
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.
 
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.
 
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 p=\frac{3a_1-a_2^2}{3}

q=\frac{9a_1a_2-27a_0-2a_2^3}{27}

Where the equation is: z^3+a_2z^2+a_1z+a_0.

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.
 
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