# Find all real values of $m$ such that all the roots of $f(x)=x^3-(m+2)x^2+(m^2+1)x-1$ are real

I have the following polynomial with real coefficients: $$f(x)=x^3-(m+2)x^2+(m^2+1)x-1$$ I have to find all real $$m$$'s so that all of the roots of $$f$$ are real.

Trying to guess a root didn't get me anywhere.

I computed $$x_1^2+x_2^2+x_3^2$$ using Vieta's relations to be $$-(m-2)^2+6$$. This has to be positive if the roots are real, so $$m\in[-\sqrt6+2, \sqrt6+2]$$.

I tried using the derivative of $$f$$ and Rolle's theorem, but the calculations get complicated quite fast. I managed to prove that m has to be somewhere in the interval $$(-\sqrt\frac32+1, \sqrt\frac32+1)$$, though I can't guarantee that this is correct. I could continue this way and I'll probably reach a solution sooner or later, but I hope there's a much more elegant solution that I've missed.

• I'm guessing the limiting cases are values $m$ that give double roots, i.e. shared roots between $f$ and its derivative. Have you tried that? Jun 14, 2020 at 14:00
• I almost missed your comment. I haven't tried that. The $m$'s that give double roots are solutions, since the other root must be real too, but I'm not sure how much would this help. Jun 14, 2020 at 14:21

I shall assume that we want three different real roots Consider $$f(x)=x^3-(m+2)x^2+(m^2+1)x-1$$ The first condition is that $$f'(x)=3x^2-2(m+2)x+(m^2+1)$$ shows two real roots which are $$x_\pm=\frac{1}{3} \left(m+2\pm\sqrt{-2 m^2+4 m+1}\right)$$ This gives the first condition $$-2 m^2+4 m+1 > 0$$

Now, you need that $$f(x_-) \times f(x_+) <0$$ that is to say $$3 m^6-4 m^5+6 m^4-22 m^3-9 m^2+26 m+23 < 0$$ which cannot be solved. Numerical calculations give $$1.558 < m < 1.756$$

• This was pretty much what I did, but I was hoping for nicer solution. Thanks anyway! Jun 14, 2020 at 14:06

hint

$$f$$ satisfies Rolle's Theorem, so if it has three roots, its derivative will have two real roots, $$\color{red}{a,b}.$$

$$f'(x)=3x^2-2(m+2)x+m^2+1$$

the reduced discriminant $$\delta=(m+2)^2-3(m^2+1)$$ $$=-2m^2+4m+1$$

should necessarily be $$>0$$.

To be sufficient, you need

$$f(a)f(b)<0$$

• This is not sufficient Jun 14, 2020 at 13:51
• This is how I obtained the last interval that I mentioned above. Unfortunately it is not enough. If the roots of the derivative are a and b, they would also have to satisfy f(a)>0 and f(b)<0. The calculations that would follow out of that are very complicated though. Edit:messed up the > and <. Jun 14, 2020 at 13:55
• The roots $x_1, x_2$ of $f'(x)=0$ should be real and be such that $f(x_1)f(x_2)<0$, in order to have a concave bump above $x$ axis followed by a convex one below $x$ axis. This can be explained regorously by invoking a properties of the so-called discriminant = resultant (f,f') if you happen to know it... Jun 14, 2020 at 13:57
• @JeanMarie Yes, i added that before reading your post. thanks quand même. Jun 14, 2020 at 14:01
• I see. I was hoping for a nicer solution, but I guess this will do. Thanks! Jun 14, 2020 at 14:05

This answer should be seen as a complement to the other answers.

A third degree curve $$y=f(x)=x^3+\cdots$$ is known to have two possible shapes

according to the fact that its derivative has sign changes (necessarily a "$$+ - +$$" pattern) or not (a simple "+" pattern), i.e., respectively two real roots $$a,b$$ or no real root. [we leave apart the limit case of one real root]. In the first case, we have a relative maximum in $$(a,f(a))$$, followed by a relative minimum in $$(b,f(b))$$. There will be 3 real roots if $$(a,f(a))$$ is above $$x$$ axis and $$(b,f(b))$$ is below $$x$$ axis ; this is equivalent to say that

$$f(a)f(b)<0$$

This condition has to be expressed in terms of parameter $$m$$.

I will use for that a method that is classical, but necessitates to know what a resultant is (explanations below). It suffices to write twice the coefficients of $$f$$, ans 3 times the coefficients of $$f'$$, with a shift for the first one and two shifts for the second one

$$Res(f,f')=\begin{vmatrix} 1& - m - 2& m^2 + 1& -1& 0\\ 0& 1& - m - 2& m^2 + 1& -1\\ 3& - 2m - 4& m^2 + 1& 0& 0\\ 0& 3& - 2m - 4& m^2 + 1& 0\\ 0& 0& 3& - 2m - 4& m^2 + 1\end{vmatrix}=0\tag{2}$$

which is identical to

$$3m^6 - 4m^5 + 6m^4 - 22m^3 - 9m^2 + 26m + 23=0$$

(the very same polynomial found by Claude).

Explanation about the calculation : the nullity of the resultant $$Res(f,g)=0$$ of 2 (parametric) polynomials $$f$$ and $$g$$ is a necessary and sufficient condition for these polynomials to have a common root ; here in the case $$g=f'$$ ; $$f$$ and $$f'$$ have a common root if and only $$f$$ has a double root. It is known to be a limit case between cases "a single real root" and "3 real roots". The sign of this resultant will change once we have crossed the case $$R(f,f')=0$$.

The resultant $$R(f,g)$$ can be computed in (at least) two ways :

• up to a factor, it is the the product of values of $$f$$ computed at the roots of $$g$$, which is formula (1)

• as the determinant computed above (in (2)) .

Remark : $$Res(f,f')$$ is called the discriminant of $$f$$ : it generalizes the discriminant of a a second degree polynomial $$ax^2+bx+c$$ for which the corresponding resultant is :

$$\begin{vmatrix} a& b& c\\ 2a& b& 0\\ 0& 2a& b\end{vmatrix}=-a(b^2 - 4ac)$$

where we recognize our classical $$b^2-4ac$$.

• Interesting. I'll look into resultants one of these days. Thanks! Jun 14, 2020 at 17:22