Consider a cubic polynomial of the form


where the coefficients are non-zero reals. Conditions for which this equation has three real simple roots are well-known. What conditions would guarantee that none of these roots is positive? In other words, what constraints on the parameters would guarantee that the polynomial has no positive roots? Please provide references also, if possible.

  • 2
    $\begingroup$ Evidently, the mathematician Sharaf al-Din Al-Muzaffar ibn Muhammad ibn Al-Muzaffar al-Tusi classified such cubics in his 12th century text Al-Mu'adalat (Treatise on Equations). In the treatise equations of degree at most three are divided into 25 different types. First al-Tusi discusses twelve types of equation of degree at most two. He then looks at eight types of cubic equation which always have a positive solution, then five types which may have no positive solution. Source: www-history.mcs.st-andrews.ac.uk/Biographies/… $\endgroup$ Sep 16 '12 at 16:23

Using Routh–Hurwitz:

  • if $a_0, a_1, a_2, a_3 > 0$, you arrive at the condition $a_2 a_1 > a_3 a_0$.
  • if $a_0, a_1, a_2, a_3 < 0$, you also arrive at the condition $a_2 a_1 > a_3 a_0$.
  • if $a_3$ and $a_2$ have different signs, then the polynomial has positive roots.
  • 2
    $\begingroup$ I would be careful with the Routh-Hurwitz criterion. For example, take the polynomial $x^2-2x+25=0$, which has no positive real roots, but fails the Routh-Hurwitz criterion $\endgroup$ Sep 18 '12 at 18:44
  • $\begingroup$ One of the requirements for constructing the Routh-Hurwitz criterion is that all the coefficients must be positive (or all negative). In this case -2x is negative so the Routh-Hurwitz method cannot be applied (See en.wikibooks.org/wiki/Control_Systems/Routh-Hurwitz_Criterion) $\endgroup$
    – rhody
    Jul 7 '15 at 0:53

You can do it like this.

If $a_3>0$ then $f(x)$ is increasing except for the region between its two turning points. To have none of the three zeros positive requires that $f(0)\geq 0$ i.e. $a_0 \geq 0$. We also need that $f(x)$ is increasing at $x=0$.

The turning points come between zeros of $f(x)$ so the other condition is that the zeros of the derivative $f'(x)$ are both negative (this deals with increasing at $x=0$ and the values can easily be expressed in terms of the coefficients). [if we did not have the condition that there were three distinct roots already, it would be possible for the derivative to have no real roots, and the condition on $a_0$ would then ensure that the one real root was not positive].

If $a_3<0$ apply the same criteria to $-f(x)$.

With $a_3>0$ the condition for three distinct real roots is equivalent to the local maximum (existing and) being greater than zero, and the local minimum being less than zero.

  • $\begingroup$ Can you comment on what happens when $a_0<0$ and $a_3>0$. Is it feasible to get all the three roots non-positive? $\endgroup$ Sep 16 '12 at 19:17
  • $\begingroup$ @transcendental The best way is to draw a graph. If $a_3>0$ then $f(x)$ is ultimately positive, increasing and unbounded. So if $a_0<0$ - which means $f(0)<0$ - there is inevitably a positive root by the intermediate value theorem [a graph will make this obvious] $\endgroup$ Sep 16 '12 at 19:29
  • $\begingroup$ I agree with all that you are saying. However, allow me to sharpen my original question. If $a_0<0$ and $a_3>0$, is it feasible to tinker with the coefficients so that $f(x)$ has no positive root? I know that the Intermediate Value Theorem (IVT) guarantees a positive root, $x_0$, in this case, but can we not use $x_0$ then to "force" $f(x)$ not to have positive roots? I was thinking along the lines of critical points of $f(x)$. Can we not use $x_0$ as part of the constraint(s) on the critical points of $f(x)$ so that $f(x)$ has no positive roots? $\endgroup$ Sep 18 '12 at 16:48

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.