Re: Reference for a cubic with a double root?

From: Ken Pledger (Ken.Pledger_at_mcs.vuw.ac.nz)
Date: 08/10/04 

Date: Tue, 10 Aug 2004 16:04:48 +1200

In article <cf97jv$r4v$1@panix2.panix.com>,
 baloglou@panix.com (George Baloglou) wrote:

> [reply address is baloglouAToswego.edu]
>
> Following the standard derivation of the formula for the solutions of a
> cubic, it is not hard to see that ax^3 + bx^2 + cx + d = 0 has a double
> root if and only if (2b^3 - 9a*b*c + 27a^2*d)^2 = 4(b^2 - 3a*c)^3. Does
> anyone know of a book where this fact is explicitly mentioned and proven?
>
> Thanks in advance,
>
> baloglouAToswego.edu

      Good old Birkhoff & MacLane, "A Survey of Modern Algebra" (2nd or
3rd edition), Chapter V, section 5.

      You may also find helpful the following reply which I made to a
related question on sci.math on the 20th January 2003.

>
> In article <f80bfed1.0301182152.68c1ef72@posting.google.com>,
> sandos_asagi@hotmail.com (Sandos) wrote:
>
> > What are the necessary and sufficient conditions for a cubic of the form
> > ax^3+bx^2+cx+d=0
> > to have three distinct roots, in terms of a,b,c,d and no
> > simplifications with substitutions?
> >
> > I take it that it's not sufficient for the discriminant to be less than
> > zero?
> > ....
>
>
> You've been given several useful replies and web sites, but I wonder
> whether you understand clearly what the discriminant of a polynomial
> actually is.
>
> If a polynomial of degree n has roots x_1, x_2, ..., x_n, then
> its discriminant is the product of n(n-1)/2 factors:
>
> (x_1 - x_2)^2.(x_1 - x_3)^2.....(x_(n-1) - x_n)^2,
>
> i.e. the product of all factors (x_i - x_j)^2 with i < j.
>
> Since this is a symmetric function of the roots, it can be expressed in
> terms of the coefficients of the original polynomial. That's what makes
> it useful.
>
> In the case n = 2 we have a quadratic, whose discriminant is just
> (x_1 - x_2)^2. It's convenient to have leading coefficient 1, so the
> quadratric is x^2 + bx + c. Then the usual proof shows that
> x_1 + x_2 = - b and (x_1)(x_2) = c. From these, you can get the
> discriminant
> (x_1 - x_2)^2 = b^2 - 4c in terms of the coefficients. Assuming b
> and c are real, you can then see that
>
> (x_1 and x_2 are real and disctinct) => (x_1 - x_2)^2 > 0
>
> => b^2 - 4c > 0,
>
> (x_1 and x_2 are equal) => (x_1 - x_2)^2 = 0
>
> => b^2 - 4c = 0,
>
> (x_1 = u + iv, x_2 = u - iv) => (x_1 - x_2)^2 = 4(i^2)(v^2) < 0
>
> => b^2 - 4c < 0,
>
> the well-known properties of the discriminant.
>
> Cubics are rather similar. This time the discriminant is
> (x_1 - x_2)^2.(x_1 - x_3)^2.(x_2 - x_3)^2. If you remove the
> second-degree term in the usual way, the cubic is x^3 + px + q = 0.
> Various proofs (ponderous or subtle) show that the discriminant
> (x_1 - x_2)^2.(x_1 - x_3)^2.(x_2 - x_3)^2 = -(4(p^3) + 27(q^2)).
> If you leave in a non-zero second-degree term, then the discriminant just
> comes out to be a more complicated expression in the coefficients.
>
> Now you should be able to work out the sign of this when x_1, x_2,
> x_3 are real and distinct, or real with at least two of them equal, or
> one real and the others complex conjugate. That will answer your question
> and (I hope) help you to understand the answer.
>
> Ken Pledger.

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