Does the Polynomial $P(x)=x^3+mx^2+nx+k$ Have Three Distinct Real Roots?

[anemone]

A polynomial $P(x)=x^3+mx^2+nx+k$ is such that $n<0$ and $mn=9k$.

Prove that the polynomial has three distinct real roots.

 

[Euge]

A polynomial $P(x)=x^3+mx^2+nx+k$ is such that $n<0$ and $mn=9k$.

Prove that the polynomial has three distinct real roots.

Spoiler

If $m = 0$, then $k = 0$ and $P(x) = x^3 + nx$. So $P$ has three real distinct roots: $0, \sqrt{-n}$ and $-\sqrt{-n}$.

Now suppose $m \neq 0$. If $m > 0$, then $P(x)$ has one sign change and $P(-x)$ has two sign changes. By Descartes' rule of signs, $P$ has one positive root and zero or two negative roots. Since $P(0) = \frac{mn}{9}$ and $P(-m) = -\frac{8mn}{9}$ have opposite signs, the intermediate value theorem ensures that $P$ has a negative root. Therefore $P$ has two negative roots. By a similar argument, if $m < 0$, then $P$ has one negative root and two positive roots. Hence in both cases, $P$ has three real roots, two of which have different signs. So it suffices to show that $P$ has no double root.

Since $P$ is cubic, if it has a double root, then that root is a critical point. The critical points of $P$ are $\frac{-m+\sqrt{m^2-3n}}{3}$ and $\frac{-m-\sqrt{m^2-3n}}{3}$. Neither of these satisfy $P$. Indeed, by the second derivative test, these points give a global min and max for $P$. But $P$ is nonconstant, so they cannot be roots of $P$. So, $P$ has no double root. Therefore, $P$ has three distinct real roots, with exactly two roots of the same sign.

 

[anemone]

Well done, Euge! And thanks for participating!

Here is one solution that I saw online that approaches it differently than you, so I will share it with you and the rest of the members:

Spoiler

Consider the derivative $P'(x)=3x^2+2mx+n$. Since $n<0$, it has two real roots $x_1$ and $x_2$. Since $P(x)\rightarrow \pm \infty$ as $x\rightarrow \pm \infty$, it is sufficient to check that $P(x_1)$ and $P(x_2)$ have different signs, i.e., $P(x_1)P(x_2)<0$.

Dividing $P(x)$ by $P'(x)$ and using the equality $mn=9k$, we find that the remainder is equal to $x\left(\dfrac{2n}{3}-\dfrac{2m^2}{9}\right)$. Now, as $x_1x_2=\dfrac{n}{3}<0$, we have $P(x_1)P(x_2)=x_1x_2\left(\dfrac{2n}{3}-\dfrac{2m^2}{9}\right)^2<0$ and we're done.