I need to solve $p(x)=aq(x)$ with multiple real $a$, where $p(x)$ and $q(x)$ are the two polynomials in $x$ (with real coefficients). The roots of $p(x)$ and $q(x)$ were found previously, i.e. these two polynomials are factored.

Is it somehow possible to solve the equation in question without multiplying $q(x)$ by $a$ and transferring everything to the left for every new $a$? Is there any workaround?

It's quite frustrating that the two polynomials $p(x)$ and $q(x)$ are known (together with their roots), but every time I substitute a new parameter value ($a$ in this case) I need to solve a totally new polynomial.

Sorry if the question is stupid. Actually I don't really believe that there's any workaround, although finding one would be really nice.

The problem is solved numerically, all polynomial roots are found as eigenvalues of the polynomial's companion matrix.

  • $\begingroup$ Sketching the zeroes' trajectories of the polynomial $p(x)-aq(x)$ could be possible using the technique called root locus, which is well known to control engineers. Unfortunately, it doesn't give explicit solutions for $a$, it "only" shows how the roots vary for $a\in [0,\,+\infty)$. Some mild restrictions on $p(x)$ and $q(x)$ also apply due to the nature of the problem (i.e. control systems). $\endgroup$ – Omicron_Persei_11 Apr 29 '14 at 10:18

Sadly no. If there was such a shortcut, solving any polynomial equation would be trivial. For particular forms of your polynomials it might be possible, but not in general.


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