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I have a cubic equation in $x$ $$x^3+bx^2+cx+d=0$$ where all the coefficients are positive.

I know that with Descartes' Rule, the equation has no positive real roots, it either has 3 negative real roots or 1 negative real root and 2 complex roots.

If I would somehow know for sure that all 3 roots are real and negative, then my problem would be solved. Is there a way of knowing if this is really the case? Or if it is not, how can I know if the complex roots have negative or positive real parts?

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4 Answers 4

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This is determined by the discriminant. You have three real roots if and only if $$b^2c^2 - 4c^3 - 4b^3d - 27d^2 + 18bcd \geq 0.$$

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    $\begingroup$ Is there a determinant form of this so it's easy to keep in mind ? $\endgroup$ Sep 2, 2023 at 3:05
  • $\begingroup$ @An_Elephant By definition, the discriminant can be written as a determinant, but for cubic polynomials this is already a $5 \times 5$ determinant, so I don't think it's particularly useful. It's more useful to use the substitution $y := x + \frac{b}{3}$ to obtain an equation of the form $y^3 + 3py + 2q = 0$. This polynomial now has only real roots if and only if $p^3 + q^2 \leq 0$. $\endgroup$
    – Klaus
    Sep 4, 2023 at 11:28
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One way is to consider the signs of $f(x) = x^3 + bx^2 + cx + d$ at the critical point(s).

$f'(x) = 3x^2 + 2bx + c$. Let the roots of $f'(x) = x_1, x_2$ and $x_1 \le x_2$.

$\therefore x_1, x_2 = \dfrac{-2b \pm \sqrt{4b^2 - 12c}}{6} \\ 4b^2 - 12c < 0 \implies x_1, x_2 \in \mathbb{C} \implies \text{ $1$ real root and $2$ complex roots} \\ f(x_1) > f(x_2) > 0 \implies \text{ $1$ real root and $2$ complex roots} \\ 0 > f(x_1) > f(x_2) \implies \text{ $1$ real root and $2$ complex roots} \\ f(x_1) \ge 0 \ge f(x_2) \implies \text{$3$ real roots}$

Note that these may not be distinct. For example, $4b^2 - 12c = 0 \text{ and } f(x_1) = 0 = f(x_2) \implies \text{ $1$ real root repeated $3$ times}$

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  • $\begingroup$ So We are calculating the critical points and using that we looking at how many times the original functions crosses the x=0 line.. isn't it? $\endgroup$ Sep 1, 2023 at 11:34
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    $\begingroup$ @PraveenKumaranP Precisely. $\endgroup$
    – Dstarred
    Sep 1, 2023 at 11:39
  • $\begingroup$ Where can I watch the reasoning... kindly give me a reference or if possible explain pls $\endgroup$ Sep 1, 2023 at 12:29
  • $\begingroup$ @PraveenKumaranP OP's cubic is positive, hence the maximum, $x_1$ is smaller than the minimum, $x_2$ (due to shape of a positive cubic). If $x_1 = x_2$ then it is a perfect cube in the form $(x - h)^3 + k$. The rest of the reasoning from here follows in the answer. $\endgroup$
    – Dstarred
    Sep 3, 2023 at 5:38
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You must look at the discriminant of the cubic equation $$\Delta=b c (b c+18 d)-\left(4 b^3 d+4 c^3+27 d^2\right)$$ If it is positive, there will be three real (negative) solutions.

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If there are two complex roots, then the sign of their real part equals the sign of $d-bc$.

Suppose the roots are $-r$ and $p\pm qi$. The cubic is

$$(x^2-2px+p^2+q^2)(x+r)\\ =x^3+(r-2p)x^2+(p^2+q^2-2rp)x+r(p^2+q^2)$$

So $d-bc$ is
$$r(p^2+q^2)-(r-2p)(p^2+q^2)+2rp(r-2p)\\ =2p(p^2+q^2+r^2-2pr)\\ =2p((p-r)^2+q^2)$$

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