How to determine the order of the real roots of a cubic equation?

Question

This is a self-answered question (I didn't find a reference, and thought of documenting this). Consider the equation $$ t^3+pt+q=0. $$ Its discriminant is $$ \Delta=-(4p^3+27q^2). $$

Suppose that it has three distinct real roots; this is equivalent to $\Delta > 0$, or $4p^3+27q^2<0$.

In particular, this forces $p<0$.

These roots can be expressed as follows : $$ x_k=2\sqrt{-\frac{p}{3}}\cos\left(\frac{1}{3}\arccos\left(\frac{3q}{2p} \sqrt{\frac{-3}{p}}\right)-k\frac{2\pi}{3}\right), $$ where $k=0,1,2$.

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Claim: $$ x_0 > x_1 > x_2. $$

How to prove this claim?

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Comment:

In order to apply $\arccos$ we must have $$-1 \le\frac{3q}{2p} \sqrt{\frac{-3}{p}} \le 1,$$

which is equivalent to $$ \frac{27q^2}{-4p^3} \le 1. $$ Since $p<0$, we can multiply this inequality by $-4p^3$, so it's equivalent to $$ 27q^2 \le -4p^3, $$ or $\Delta \ge 0$.

Answer 1

The three numbers $x_k=t\cos\left(\theta-k\frac{2\pi}3\right)$ ($k=0,1,2$) are supposed to be distinct and $t$ is positive.

Moreover, since $\theta\in[0,\pi/3],$ we have $$\cos\theta-\cos\left(\theta-\frac{2\pi}3\right)=-2\sin\left(\theta-\frac\pi3\right)\sin\frac\pi3\ge0$$ and $$\cos\left(\theta-\frac{2\pi}3\right)-\cos\left(\theta-\frac{4\pi}3\right)=-2\sin(\theta-\pi)\sin\frac\pi3\ge0.$$ Therefore, $$x_0>x_1>x_2.$$

Answer 2

Set $$ m:=\frac{1}{3}\arccos\left(\frac{3q}{2p} \sqrt{\frac{-3}{p}}\right), $$ we have $0 \le m \le \pi/3$. (since the range of $\arccos$ is $[0,\pi]$).

The analysis in the question shows that $\Delta >0$ if and only if $-1<\frac{3q}{2p} \sqrt{\frac{-3}{p}}<1$, which is equivalent to the strict inequality $0<m<\pi/3$.

Then, up to a multiplication by the positive constant of $2\sqrt{-\frac{p}{3}}$, we have $$ x_k=\cos\left(m-k\frac{2\pi}{3}\right), $$ or $$ x_0=\cos(m), x_1=\cos(m-\frac{2\pi}{3}),x_2=\cos(m-\frac{4\pi}{3}). $$ Thus, $x_0 > x_1 > x_2$ is equivalent to $$ \cos(m) >\cos(m-\frac{2\pi}{3})>\cos(m-\frac{4\pi}{3}), $$ for any $m \in (0,\pi/3)$.

This can by differentiating the inequality $\cos(x) >\cos(x-\frac{2\pi}{3})$ and observing that the difference is monotone on $(0,\pi/3)$, and similarly for the second inequality.