How is the resultant obtained from a pair of quadratic polynomials? I'm reading about resultant theory from Peter Stiller's notes (Link to pdf). The author mentions that, given two univariate polynomials
$$f(x)=a_r x^r+ ...+a_1 x+ a_0,
\qquad g(x)=b_s x^s+ ...+b_1 x+ b_0,$$
there are common roots iff their resultant $R_{r,s}(f,g)$ is zero, where the resultant is defined as the determinant of the Sylvester matrix of the two polynomials.
For example, for $r=s=2$, the resultant reads
$$R_{2,2}(f,g) = 
a_0^2 b_2^2 + a_0 a_2 b_1^2 - a_0 a_1 b_1 b_2 + a_1^2 b_0 b_2 - a_1 a_2 b_0 b_1 + a_2^2 b_0^2 - 2 a_0 a_2 b_0 b_2.$$
I'm trying to get a better understanding of where this expression comes from.
I understand that the resultant is what should come out imposing the two polynomials to have common roots. However, if I simply multiply $f(x)$ by $b_0/a_0$ (assuming $a_0\neq 0$) and remove the corresponding common term in both polynomials, I get a first order polynomial in $x$. Similarly if I remove the $x^2$ term from the two polynomials.
From these first-order polynomials, I can easily deduce the solutions to the system. How would one combine these solutions to obtain the expression for the resultant?

Consider two quadratic polynomials $a(x)=a_0 + a_1 x + a_2 x^2$ and $b(x)=b_0 + b_1 x + b_2 x^2$. Performing the Euclidean algorithm we get,
$$a(x) = q_0 b(x) + (r_0 + r_1 x), 
\qquad q_0 = \frac{a_2}{b_2}, \\
r_0 = a_0 - q_0 b_0 = a_0 - \frac{a_2}{b_2} b_0,
\qquad r_1 = a_1 - q_0 b_1 = a_1 - \frac{a_2}{b_2}b_1.$$
Dividing $b(x)$ by $r(x)$ we then get
$$b(x) = (q_0' + q_1' x) (r_0 + r_1 x) + r_0',
\qquad q_1' = \frac{b_2}{r_1} = \frac{b_2^2}{a_1 b_2 - a_2 b_1}, \\
q_0' = \frac{1}{r_1}\left[
b_1 - q_1' r_0
\right] = \frac{1}{r_1}\left[
b_1 - q_1' r_0
\right],
\qquad r_0' = b_0 - q_0' r_0.$$
If $a$ and $b$ have a common root, then we need $r_0'=0$, as each remainder must be a multiple of the GCD of the polynomials. Thus we get the condition
$$r_0' = b_0 - q_0' r_0 = 0.$$
Replacing the known expressions for $q_0'$ and $r_0$ we do indeed get a condition on the coefficients $a_i, b_i$, but this hardly seems like the best approach to the problem. I don't see how the Sylvester matrix is supposed to arise, and it doesn't seem prone to easy generalisation.
 A: 
I'm trying to get a better understanding of where this expression [the resultant] comes from

For the case of two quadratics, a common root means the following equations are satisfied for at least one value of $x$. The system includes the two original equations, plus the same equations multiplied by $x$.
$$
\begin{cases}
  a_2 x^3 + & a_1 x^2 + & a_0 x && = 0 \\
& a_2 x^2 + & a_1 x   + & a_0 & = 0 \\
  b_2 x^3 + & b_1 x^2 + & b_0 x && = 0 \\
& b_2 x^2 + & b_1 x   + & b_0 & = 0 \\
\end{cases}
$$
Consider this as a system of linear equations in $x_0 = 1, x_1 = x, x_2 = x^2, x_3 = x^3$.
$$
\begin{cases}
  a_2 \cdot x_3 & +\; a_1 \cdot x_2 & +\; a_0 \cdot x_1 + & \,0 \cdot x_0 & = 0 \\
  \,0 \cdot x_3 & +\; a_2 \cdot x_2 & +\; a_1 \cdot x_1   + & a_0 \cdot x_0 & = 0 \\
  b_2 \cdot x_3 & +\; b_1 \cdot x_2 & +\; b_0 \cdot x_1 + & \,0 \cdot x_0 & = 0 \\
\,0 \cdot x_3 & +\; b_2 \cdot x_2 & +\; b_1 \cdot x_1   + & b_0 \cdot x_0 & = 0 \\
\end{cases}
$$
This is a linear homogeneous system which has the non-trivial solution $(1, x, x^2, x^3)$ so its matrix must be singular, which happens to be precisely the Sylvester matrix. This amounts to the following determinant being zero.
$$
\left| 
\begin{matrix}
\;a_2 \;&\; a_1 \;&\; a_0 \;&\; 0 \\
\;0 \;&\; a_2 \;&\; a_1 \;&\; a_0 \\
\;b_2 \;&\; b_1 \;&\; b_0 \;&\; 0 \\
\;0 \;&\; b_2 \;&\; b_1 \;&\; b_0 \\
\end{matrix}
\right| \;\;=\;\; 0
$$
Expanding the determinant results in the expression posted for $R_{2,2}(f,g)$.
In the general case of polynomials $f$ of degree $r$ and $g$ of degree $s$, multiplying the equation $f(x) = 0$ by $x, x^2, \dots, x^{s-1}$ and $g(x) = 0$ by $x, x^2, \dots, x^{r-1}$ gives a system of $r+s$ equations with a similar Sylvester matrix.

Performing the Euclidean algorithm we get [...]

Looking at it from another angle, the first step of the euclidean division eliminates the $x^2$ term between equations, then the second step eliminates the $x$ term. In the end, this is solving the same equations by successive elimination, instead of determinants, but it becomes more laborious to do for higher degrees.
