Re: Reference for a cubic with a double root?

From: George Baloglou (baloglou_at_panix.com)
Date: 08/12/04 

Date: 12 Aug 2004 17:50:42 -0400

In article <cfcf5h$9em$1@panix1.panix.com>, I wrote:

>[reply address is baloglouAToswego.edu]
>
>[snip]
>
>(3) A 'Calculus' approach according to which there will be a double root
>iff one of the derivative's root(s), that is -b/3a +- sqrt(b^2-3ac)/3a, is
>also a root of the cubic: substituting into the cubic, we see that all the
>radicals are eliminated -- regardless of whether it's the + or - sign that
 ^^^^^^^^^^^^^^^^^^^^^^^
To be more precise, we end up with +- 2(b^2 - 3ac)*sqrt(b^2 - 3ac) =
2b^3 - 9abc + 27(a^2)d , from which the condition right below follows.

>is in effect -- and that we obtain a value of zero for the cubic precisely
>when (2b^3 - 9abc + 27(a^2)d)^2 = 4(b^2 - 3ac)^3. [This is 'my' way.I
      ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
Notice that we end up with a triple root if and only if both sides are zero.

>[snip]
> baloglouAToswego.edu
>
>One possible reason for the creature's sudden fit of fury may have been an
>unconfirmed report that it was "kicked by somebody in business class" on
>its way through the cabin [http://news.bbc.co.uk/2/hi/europe/3551672.stm]

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