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I'm working on an algorithm to calculate all real roots of a quartic equation.

At the moment I have a solution using the method of Descartes and Vietas substitution. My approach is:

  1. Normalize quartic
  2. Convert to depressed quartic
  3. Decompose into quadratic factors $\rightarrow$ yields resolvent cubic
  4. Convert resolvent cubic to depressed cubic
  5. Solve the depressed cubic (which involves calculation of a cube root)
  6. Reverse further previous substitutions (which involves calculation of square roots)

At the moment I'm assuming that all variables are real. If a complex number would occur (e.g. while calculating the cube root or square roots) I'm neglecting the solution at further calculations. However, I'm not sure if there could be the possibility that one of my complex immediate results could lead to a real root of the quartic.

My question: Is this possible? Or in other words: Would it be necessary to also consider complex immediate results during the calculations?

If the answer would be yes, than my second question is: Is there (another) method that allows solving the quartic equation without the need for operating with complex numbers?

theNewOne
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    what if the quartic equation has complex roots? – J. W. Tanner Jan 09 '20 at 17:32
  • If we consider, say, Cardano's formula for cubic roots, it should be entirely possible that complex elements show up midway through the calculation that end up canceling, thus resulting in a real number in the end. That said, a polynomial with real coefficients may have complex roots.ca – Fimpellizzeri Jan 09 '20 at 17:42
  • How do you do step 3? If it means what you said, the solution is obvious. – herb steinberg Jan 09 '20 at 17:42
  • @J.W.Tanner I'm not interessted in the complex roots. – theNewOne Jan 09 '20 at 18:01
  • @Fimpellizieri "If we consider, say, Cardano's formula for cubic roots, it should be entirely possible that complex elements show up midway through the calculation that end up canceling, thus resulting in a real number in the end." That's excactly what I'm asking. Is it really possible that this could happen? – theNewOne Jan 09 '20 at 18:02
  • Yes, it is. ${}$ – Fimpellizzeri Jan 09 '20 at 18:05
  • @Fimpellizieri Ok. Do you also have an answer to my second question than? (regarding the solving using only real numbers) – theNewOne Jan 09 '20 at 18:07
  • I would think the answer is no, since quartics may have complex roots. – Fimpellizzeri Jan 09 '20 at 18:12
  • @herbsteinberg I do it like it is shown here: https://math.stackexchange.com/a/1039289/572652 Could you be a bit more specific regarding what is "obvious" please? – theNewOne Jan 09 '20 at 18:15
  • Step 3 is the hard part. Once you have two quadratics, each can be solved by the usual quadratic formula. – herb steinberg Jan 09 '20 at 18:26
  • @herbsteinberg My question is not "how to do it". My question is, whether I have to consider all complex iintermediate results to find the real roots when using the procedure that I listed. – theNewOne Jan 09 '20 at 19:50

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If the roots are irrational, there are cases where solution "by radicals" will inevitably involve complex numbers. However, complex numbers can be avoided by using trigonometric functions. See https://en.wikipedia.org/wiki/Casus_irreducibilis for details.

(Furthermore, it is always possible to find the real roots of a polynomial through successive approximation techniques. While this sounds like "cheating", in fact, irrational numbers, such as some radicals and trig values, are in fact found by successive approximation. The question is really whether one only allows certain "canned", or "well known", approximations such as roots and/or trig functions, or whether one also allows "custom" approximations.)

Math Keeps Me Busy
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