Eliminating a variable with a resultant

Consider the original pair of polynomial equationsTreat them as polynomials in $y$ with coefficients in $\mathbb{R}[x]$. Their Sylvester matrix and resultant areEvery common solution must therefore have $x=-1$ or $x=1$. Back-substitution into $F=0$ gives $y=x^2=1$, and both pairs also satisfy $G=0$. The complete real solution set isHere elimination and direct substitution agree: subtracting $F$ from $G$ gives $2x^2-2$, exactly the same condition.

In general, first solve the one-variable resultant equation, then recover and check the other coordinate in both original polynomial equations. A real coordinate candidate can correspond only to nonreal values of the eliminated variable. Also, vanishing leading coefficients require the separate check in degree drop in a specialized resultant. The general elimination-and-back-substitution method is described in the lecture's §3.1.^[ref]

The actual-degree condition in the resultant theorem matters when coefficients depend on a parameter. TakeThe nominal degree-one Sylvester matrix is $\left(\begin{array}{cc}t& 1\\ t& 2\end{array}\right)$, whose determinant is $t$. Its specialization at $t=0$ therefore vanishes. But the specialized polynomials are the nonzero constants $1$ and $2$, which have no common root at all.

For $t\ne 0$, the roots of a polynomial are $-1/t$ and $-2/t$; neither has a finite limit as $t\to 0$. Both leading coefficients vanish at the exceptional parameter. The fixed-size determinant has retained the original degree format even though the specialized polynomial degrees have dropped. Recompute using the actual specialized polynomials and check the original equations before accepting a zero of a parameter-dependent resultant as a solution.