Date: Fri, 23 Apr 1999 09:47:13 +0200 (METDST)
Subject: Question about 3rd degree polynomials
What is the general solution to the equation with the form:
a*x^3 + b*x^2 + c*x + d = 0
I have once seen a solution to this a few years ago, but I do not recall if it was a general solution. What I do know, is that you could simplify this equation to:
a*x'^3 + p*x' + q = 0
where x' = x - k, and k = -b/3a. p = 3*a*k^2 + 2*b*k + c and q = k^3 + b*k^2 + c*k + d
You can derive for what values of "a", "p", and "q" the equation has one or three (real) solutions. But I do not know how to proceed in order to find these solutions.
I would be very grateful if you could let me know the answer to this.
for a real number x, make the substitutions x' = x - k and k = -b/3a. In doing so, the original cubic equation now becomes
where p = 3ak2 + 2bk + c and q = k3 + bk2 + ck + d
for x', one can use the method (formula) of Niccolo Fontano [or Tartaglia] (1499-1557), which was publicized in the book "Ars Magna" by Gerolamo Cardano (1501-1576). Clearly once x' is found, then the solution x to the original cubic is x = x' - k.
The solution x' to
is given by
Because both u and v require the calculation of the square root of q2/4 + p3/27, the formula is most easily applicable to situations in which q2/4 + p3/27 is greater than or equal to 0.
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