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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.

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  • $\begingroup$ 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
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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.

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  • $\begingroup$ 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
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    $\begingroup$ Your solution doesn't hold in general. How do you get it? $\endgroup$ – Fabio Lucchini Oct 14 '18 at 11:47

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