How can I show, that if $u:= ax + b$ and $v:= x^2 + cx + d$ , $(a, b, c, d ∈ \mathbb{R})$ polinoms have common root, if and only if $det(A[u, v]) = 0$, where $A[u, v] = \begin{bmatrix} a & b & 0\\ 0 & a & b\\ 1 & c & d \end{bmatrix}$ .
Should I do this with resultant? And how? I'm looking for guidance, not the exact solution.
On
$u$ and $v$ have a common root if and only if there are non-zero polynomials $\alpha$, with $\deg \alpha \leq \deg u -1$, and $\beta $, with $\deg \beta \leq \deg v -1$, such that
$$ \beta(x) u(x) + \alpha(x) v(x) = 0. \; \; (1)$$
(We can see this by denoting the common factor by $t$, non-constant polynomial, such that $u = tu_1$ and $v=tv_1$. We can then choose $\beta = v_1$ and $\alpha = -u_1$.)
In our example, $\alpha(x) = e$ and $\beta(x) = fx+g$, respectively.
The null polynomial $(1)$ can be rewritten as
$$ (fa+e)x^2 + (fb+ga +ec)x + (gb +ed) = 0, $$ which is equivalent to the system of equations in unknowns $ f, g, e$ $$ a\cdot f + 0\cdot g + 1\cdot e =0 $$ $$ b\cdot f + a \cdot g + c\cdot e = 0 $$ $$ 0\cdot f + b\cdot g + d \cdot e = 0 $$ Solution $ f, g, e$ is non-trivial if and only if $\det A = 0$.
Brute force solution: For $a\not= 0$, the common root can only be
$$-\frac{b}{a}. $$
In this case:
$$ v\left(-\frac{b}{a} \right) = 0,$$
which is equivalent to:
$$ b^2 -cba +da^2 =0.$$
One can then verify that: $$\det A = b^2 -cba +da^2. $$
$\det A[u,v]=0$ if and only if there exists a nonzero vector $X$ such that $$\begin{pmatrix} a & b & 0\\ 0 & a & b\\ 1 & c & d \end{pmatrix} X =0.$$ If $\alpha$ is a common root, then $X=\begin{pmatrix} \alpha^2\\ \alpha \\ 1\\ \end{pmatrix}$ works. Now if such an $X$ exists can you prove that there is a common root?