# How to tell if a cubic equation with positive coefficients has three real, negative roots

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?

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

• Is there a determinant form of this so it's easy to keep in mind ? Sep 2, 2023 at 3:05
• @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 $p^3 + q^2 \leq 0$. Sep 4, 2023 at 11:28

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, $$4b^2 - 12c = 0 \text{ and } f(x_1) = 0 = f(x_2) \implies \text{ 1 real root repeated 3 times}$$

• 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? Sep 1, 2023 at 11:34
• @PraveenKumaranP Precisely. Sep 1, 2023 at 11:39
• Where can I watch the reasoning... kindly give me a reference or if possible explain pls Sep 1, 2023 at 12:29
• @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. Sep 3, 2023 at 5:38

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.

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
$$r(p^2+q^2)-(r-2p)(p^2+q^2)+2rp(r-2p)\\ =2p(p^2+q^2+r^2-2pr)\\ =2p((p-r)^2+q^2)$$