On the number of roots of the polynomial $x^3+Ax^2+1=0$

I have the following cubic equation

$$x^{3}+Ax^{2}+1=0$$

where $$A$$ is an arbitrary (real) number.

I know that either:

• The 3 roots will be real.
• One root will be real and the other two will be complex conjugates of each other.

I would like to find out

• For what value/values of A the roots change from 3 real roots to one real and two complex roots.
• The signs of each of the real roots (both when they are all real and when there is only one real root)

Is there an analytical way of finding this as a function of $$A$$ or the only option is to solve the cubic numerically?

• Google Cardano's method. – hamam_Abdallah Jan 6 at 19:44

If we can show that the value of $$A$$ changes the amount of maxima or minima from 1 to 2 such that one extreme is at a positive $$y$$-location and the other at a zero or negative $$y$$-location, then we know that this value of $$A$$ is the border for all real solutions versus 1 real and 2 complex solutions.

Denote $$f(x) = x^3 + Ax^2 + 1.$$

$$f'(x) = 3x^2+2Ax=x(3x+2A) = 0.$$

We see that $$f(0)$$ is always either a local maximum or minimum and $$f(0) > 0.$$ The other zero is given by $$x = -2A/3 = \xi.$$

We need $$f(\xi) \leq 0.$$

$$f(\xi) = \frac{4}{27}A^3 + 1 \leq 0$$

solves for the values of $$A$$ where 3 real solutions are guaranteed.

If there are 3 real roots, then since $$x = 0$$ is always a local extreme, at least one has to be positive and another has to be negative. Also, if $$A \leq -3/\sqrt[3]{4}$$ and the location of the moving extreme is $$x=-2A/3 > 0,$$ then the remaining real root has to be positive.

If there's only 1 real root, then $$A > -3/\sqrt[3]{4}$$ and the moving extreme is located at $$x < \sqrt[3]{2}.$$ Thus the only real root must be negative.

• How do we find the signs of the real roots? By actually finding them or is there a shortcut? – JennyToy Jan 6 at 20:21
• @JennyToy I should've reread the questions before submitting the first time, sorry! How's the edit? – AEngineer Jan 6 at 20:45

hint

Put $$A=-\frac 32B$$ and $$f(x)=x^3-\frac 32Bx^2+1.$$

$$f'(x)=3x(x-B).$$

$$f(0)=1$$ to have three real roots, we need

$$B>0 \text{ and } f(B)<0.$$

• To find when there are 3 real roots or one only use the sign of the discriminant $\Delta = -4A^{3}-27$. Thus if $A=-3/2^{2/3}$ there are 3 equal real roots. If $A<-3/2^{2/3}$ there are 3 real distinct roots. And if $A>-3/2^{2/3}$ there is one real root and two (complex conjugate) complex roots. Unfortunately to find the signs of the real roots we probably need to actually find the roots. – user2175783 Jan 6 at 20:13
• @user2175783 If $A=-\frac{3}{2^{2/3}}$, then there are a negative zero and a double zero on the positive axis. – Hanno Jan 12 at 22:17

The stationary points of $$f(x)$$ lie at $$x=0$$ and at $$x=-\frac{2A}{3}$$ and the sign of $$f(0)f\left(-\tfrac{2A}{3}\right) = 1+\tfrac{4}{27}A^3$$ decides if there are three real roots (<0), a simple real root (>0) or a double real root and a simple real root (=0). This happens since the sign of $$f(0)f\left(-\tfrac{2A}{3}\right)$$ only depends on the stationary values having the same sign or not.

Too late at the party. But it looks good, so let's enter anyway.

TL;DR $$\quad f(x)=x^3+Ax^2+1\,$$ always has a simple zero on the negative axis. The other two zeros are

• real, positive and distinct, if $$\,A<-1.88988\:=\:-\dfrac{3}{\sqrt[3]{4}}\:=\:A_1$$
• merging into a double zero at $$x=\sqrt[3]{2}$$, if $$\,A=A_1$$
• non-real (thus complex-conjugate), if $$\,A>A_1$$

The party-giver also asked if there is an analytical way of finding this as a function of $$A$$?
Let's take a step in this direction, concentrating on the case of "one real zero", and express the negative zero as a function of $$A$$:
Let $$\,d=\big(\frac A3\big)^3 + \big(\frac 12\big)^2$$, and assume $$d>0$$ which corresponds to $$\,A>A_1$$. Define $$r \:=\:\left(-\frac 12 -\sqrt d\right)^\frac 13 +\, \left(-\frac 12 +\sqrt d\right)^\frac 13\,,$$ $$r\,$$ is observed to be strictly negative. Then \begin{align} r^2\:= & \;\left(-\frac 12 -\sqrt d\right)^\frac 23 +\, \left(-\frac 12 +\sqrt d\right)^\frac 23 -\frac23 A \\[2ex] r^3\:= & \; -\frac23 Ar + \left(-\frac 12 -\sqrt d\right) +\, \left(-\frac 12 +\sqrt d\right) \\ & \quad -\frac13 A \left\{ \left(-\frac 12 -\sqrt d\right)^\frac 13 +\, \left(-\frac 12 +\sqrt d\right)^\frac 13\right\} \\ = & -Ar - 1 \end{align} Whence $$1+Ar+r^3=0\quad\Longleftrightarrow\quad\frac1{r^3}+A\frac1{r^2}+1 =0 \\[2ex] \Longrightarrow\quad x \:=\: \frac1r \:=\: -r^2-A \:=\: -\frac13 A -\left(\sqrt d + \frac 12\right)^\frac 23 -\, \left(\sqrt d -\frac12\right)^\frac 23$$ satisfies $$x^3+Ax^2+1=0$$.

Added in Edit: Driven by user2175783's comment as of below, I found Lecture 4 in Mathematical Omnibus: 30 Lectures on Classic Maths by D. Fuchs + S. Tabachnikov: It gives an extensive and readable account of how to explicitly solve the cubic and quartic equations.

Visual TL;DR

• Wouldnt it be possible to generalize the algebra for the negative root for $A<A_{1}$? – user2175783 Jan 13 at 0:14
• @user2175783 I did not look closer at that case. I'd say there's no ambiguity for the sign of $\sqrt d$ because $r$ is a symmetrical expression. But afterwards complex cube roots have to be considered which might involve some choice. – Hanno Jan 13 at 9:31

The discriminant of a cubic polynomial with real coefficients is $$0$$ when the polynomial has two (or more) equal roots, positive when there are three distinct real roots, and negative when there are one real and two non-real roots. In your case it is $$-4 A^3 - 27$$.