Quartic Quasi-Discriminant

Question

I'm trying to find a condition on a, b and c for the quartic $P(x)=x^4+ax^3+bx+c$ to have a triple root.

Using the Multiple Root Theorem, it's easy to show that if it has a triple root, it must be $\alpha=-\frac{a}{2}$.

So the usual method of finding the condition is to substitute this back into the original polynomial. In other words, $P\left ( -\frac{a}{2} \right )=0$ should yield the (necessary?) condition for the polynomial to have a triple root.

After some computation, I get

• $a^4+8ab=16c \qquad (1)$

Here is where I come across issues.

In order to check that this actually works, I set $a=1, b=1$ and found $c=\frac{9}{16}$, which does NOT have a triple root from the graphing calculator.

My thought process then was "Oh! It must be because all triple roots must also be double roots as well, so perhaps I need to substitute $x=-\frac{a}{2}$ back into $P'(x)$ to get a second condition that must be satisfied!"

So I worked with

$P'\left ( -\frac{a}{2} \right )=0 $ to get

• $a^3+4b=0 \qquad (2)$

My thought process afterwards was "Okay, so if I find a,b,c satisfying both (1) AND (2), then that should yield a polynomial with a triple root!"

I set $a=2$, which gave me $b=-2$ using (2) which then using (1) yielded $c=-1$. So this triplet should satisfy both conditions (1) and (2).

But the graph does not have a triple root unfortunately.

What is going on? Is it perhaps to do with the fact that the conditions are necessary, but not sufficient?

Answer 1

You have a triple root at $x$ when $P(x)=0\land P'(x)=0\land P''(x)$ simultaneously.

$$x^4+ax^3+bx+c=0\\4x^3+3ax^2+b=0\\12x^2+6ax=0.$$

An obvious solution is $x=0$, with $b=c=0$ and arbitrary $a$. There is another for $x=-\dfrac a2$.

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$$x=-\frac a2,b=-\frac{a^3}4,c=-\frac{a^4}{16}.$$

Answer 2

The case of a root that is at least double is well-known and characterized by the anihilation of the discriminant.

Let us consider the specific case where there is exactly a triple root $t$, the other root $s$ being simple. $$x^4+ax^3+bx+c = (x-s)(x-t)^3\ \ \text{for certain} \ s, t, \ s \neq t \tag{1}$$

This case is amenable to something rather interesting .

Expanding the RHS of (1) and identifying similar coefficients, we obtain the system :

$$\begin{cases}a + 3t + s&=&0 \ \ \ \ \ (a)\\ 3t(t+s)&=&0 \ \ \ \ \ (b)\\ t^3 +3st^2 + b&=&0 \ \ \ \ \ (c)\\ st^3 - c&=&0 \ \ \ \ \ (d) \end{cases}$$

Condition (b) permit to split the issue into two subcases :

• If $t=0$, one obtains $b=c=0$ and $s=-a$, whence the first family of solutions, polynomials

$$P_a(x)=x^4-ax^3=x^3(x-a)$$

which indeed have $0$ as a triple root and $a \neq 0$ as the simple root.

• If $t \neq 0$, condition (b) gives $s=-t$. Plugging this expression of $s$ into (a),(c) and (d) finally gives the second family of solutions :

$$Q_t(x)=x^4-2tx^3+2t^3x-t^4=(x+t)(x-t)^3$$

once again depending on a single parameter ($t \neq 0$).

Here is a graphical representation of some polynomials $Q_t$ for $t=-4...4 \ \ (t \neq 0)$ :

Fig. 1. Graphical representation of different polynomials $Q_t$. For $t = -2$, one gets the curve in dark red with roots in $t=-2$ (triple root) and in $s=2$ (simple root).