Date: Fri, 23 Apr 1999 09:47:13 +0200 (METDST) Sender: Patrick Subject: Question about 3rd degree polynomials
Dear Sir/Madam, 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.
Sincerely,
Hi Patrick ^{3} + bx^{2} + cx + d = 0for a real number x, make the substitutions x' = x - k and k = -b/3a. In doing so, the original cubic equation now becomes ^{3} + px' + q = 0.where p = 3ak ^{2} + 2bk + c and q = k^{3} + bk^{2} + ck + d
To solve ^{3} + px' + q = 0.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 ^{3} + px' + q = 0.is given by ^{2}/4 + p^{3}/27)^{1/2}]^{1/3}and ^{2}/4 + p^{3}/27)^{1/2}]^{1/3}Because both u and v require the calculation of the square root of q ^{2}/4 + p^{3}/27, the formula is most easily applicable to situations
in which q^{2}/4 + p^{3}/27 is greater than or equal to 0.Doug
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