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I am trying to find a relation between coefficients $a$ and $b$ of the equation $a x^3 + b x^2 - x + 2 =0$ so that I get positive real roots of the equation (i.e. $x\geq0$). Any help on how to do this on Mathematica 12?

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    $\begingroup$ Check the function Discriminant. $\endgroup$ Jul 10, 2023 at 13:39
  • $\begingroup$ @DanielLichtblau Thanks I have taken the discriminant, g[a_, b_] := Discriminant[a^2 x^3 + 2 b x^2 - x + 2, x] and done a region plot, RegionPlot[g[a, b] >= 0, {a, -2, 2}, {b, -2, 2}, FrameLabel -> {"a", "b"}]. It solves my problem! $\endgroup$
    – misphyz
    Jul 10, 2023 at 14:27

1 Answer 1

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Try:

 Solve[a*x^3 + b*x^2 - x + 2 == 0 && x >= 0, x, Reals]

You should get the roots (3) with the conditional domains for the coefficients. The notebook will return them as icons which you can expand and examine.

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  • $\begingroup$ +1 You could also append //ToRadicals or //ToRadicals//InputForm $\endgroup$
    – Bob Hanlon
    Jul 10, 2023 at 14:30
  • $\begingroup$ Using ToRadicals will not readily help due to the casus irreducibilis. $\endgroup$ Jul 10, 2023 at 17:43
  • $\begingroup$ @DanielLichtblau - ToRadicals alleviates the panic that some experience when encountering Root expressions. In this case, the order of the polynomial is low enough that ToRadicals will always work to eliminate the Root representation. $\endgroup$
    – Bob Hanlon
    Jul 10, 2023 at 18:57
  • $\begingroup$ I realize it will eliminate the Root expressions. The problem is that both cases of one and three real roots will be indistinguishable (ignoring the case where a=0 and there are only two roots). That's the gist of the casus irreducibilis: the radical expressions will have explicit imaginary values even when they are real-valued.... $\endgroup$ Jul 10, 2023 at 21:23
  • $\begingroup$ ...Example: In[2702]:= rads = ToRadicals[SolveValues[x^3 + -5*x^2 - x + 2 == 0, x]] N[rads] // Chop Out[2702]= {5/3 - (14 (1 + I Sqrt[3]))/( 3 (1/2 (241 + 9 I Sqrt[367]))^(1/3)) - 1/6 (1 - I Sqrt[3]) (1/2 (241 + 9 I Sqrt[367]))^(1/3), 5/3 - (14 (1 - I Sqrt[3]))/(3 (1/2 (241 + 9 I Sqrt[367]))^(1/3)) - 1/6 (1 + I Sqrt[3]) (1/2 (241 + 9 I Sqrt[367]))^(1/3), 1/3 (5 + 28/(1/2 (241 + 9 I Sqrt[367]))^( 1/3) + (1/2 (241 + 9 I Sqrt[367]))^(1/3))} Out[2703]= {-0.6874, 0.568373, 5.11903} $\endgroup$ Jul 10, 2023 at 21:23

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