Respected all.

I am willing to know if my following computation is correct or not. I request you to please give me your suggestion how to improve in case any error is found.

The question is : If $a_1 x^3+b_1x^2+c_1x+d_1=0$ and $a_2x^3+b_2x^2+c_2x+d_2=0$ be two cubic polynomials with real coefficients and they have a common root say $x=m\in \mathbb{R}$, how to determine it ?

My try:

Since $x=m$ is a common root so we must have \begin{align*} &a_1m^3+b_1m^2+c_1m+d_1=0\\ &a_2m^3+b_2m^2+c_2m+d_2=0 \end{align*} Hence we must have: $$\frac{m^3}{b_1c_2-b_2c_1}=\frac{m^2}{c_1d_2-c_2d_1}=\frac{m}{d_1a_2-d_2a_1}=\frac{1}{a_1b_2-a_2b_1}$$

And from here, we get the value of $m$.

Please tell me if it is correct. In case if not, please correct it through editing.

Thank you in advance

p.S. I know about Sylvester matrix which enables us to verify if two given polynomials of degree $m,n$ will share a common root by the Sylvester matrix which determinant is called the Resultant of those two polynomials. But I do not know how to use Sylvester matrix to determine that common root. Hence I tried the above method.

Please let me know your suggestion.

  • $\begingroup$ Note that $m$ is also a root of the gcd of the two polynomials. That gcd can be found with euclidean algorithm and has degree less than 3 (if the two cubic are not equivalent) $\endgroup$ – Fabio Lucchini Oct 13 '18 at 22:31
  • $\begingroup$ It is easy enough to create two real cubic polynomials with a common real root. Have you tried some examples? Do they all satisfy your proposed equation? $\endgroup$ – David K Oct 14 '18 at 0:06

By putting \begin{align*} &f(m)=a_1m^3+b_1m^2+c_1m+d_1=0\\ &g(m)=a_2m^3+b_2m^2+c_2m+d_2=0 \end{align*} we get $$a_2f(m)-a_1g(m)=(a_2b_1-a_1b_2)m^2+(a_2c_1-a_1c_2)m+(a_2d_1-a_1d_2)=0$$ hence if $a_2b_1-a_1b_2\neq 0$ then $$m=\frac{-(a_2c_1-a_1c_2)\pm\sqrt{(a_2c_1-a_1c_2)^2-4(a_2b_1-a_1b_2)(a_2d_1-a_1d_2)}}{2(a_2b_1-a_1b_2)}$$ otherwise $$m=-\frac{a_2d_1-a_1d_2}{a_2c_1-a_1c_2}$$

Note that if $a_2b_1-a_1b_2=a_2c_1-a_1c_2=0$, then also $a_2d_1-a_1d_2=0$, hence $a_2f=a_1g$, that's $f$ and $g$ have the same roots.

  • $\begingroup$ what if $a_2b_1-a_1b_2=0$ ? $\endgroup$ – Anjan3 Oct 13 '18 at 22:40
  • $\begingroup$ Nice solution indeed. But it is still not clear to me whether my solution is correct or not. :-( $\endgroup$ – Anjan3 Oct 14 '18 at 11:45
  • 1
    $\begingroup$ Your solution doesn't hold in general. How do you get it? $\endgroup$ – Fabio Lucchini Oct 14 '18 at 11:47

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.