For a coprime pencil of degree 2 in characteristic zero, q, p and the Wronskian p′q − pq′ span the polynomials of degree at most 2

Let $K$ be a field of characteristic zero, let $p, q \in K[X]$ be coprime with $\max(\deg p, \deg q) = 2$, and let $J = p'q - pq'$ be their Wronskian. Then every $R \in K[X]$ of degree at most $2$ is $R = aq + bp + cJ$ for some $a, b, c \in K$.

Characteristic zero is needed: in characteristic $2$, $p = X^2$ and $q = 1$ give $J = 0$.

Proof. Write $p_i, q_i$ for the coefficients of $X^i$ and $[ij] = p_iq_j - p_jq_i$. By DiazModulus.quadratic_pencil_wronskian_identities, $J = [21]X^2 + 2[20]X + [10]$, so the matrix $M$ whose rows are the coefficient vectors of $q$, $p$ and $J$ in the basis $(1, X, X^2)$ has determinant $-2\,\mathrm{Res}$, with $\mathrm{Res} = [20]^2 - [21][10]$. By DiazModulus.quadratic_pencil_resultant_ne_zero, $\mathrm{Res} \neq 0$, and $2 \neq 0$ in characteristic zero, so $M$ is invertible. If $r$ is the coefficient vector of $R$, the vector $v = rM^{-1}$ satisfies $vM = r$, which says, coefficient by coefficient, that $R = v_0q + v_1p + v_2J$; the coefficients of degree at least $3$ vanish on both sides.

Novelty. Classical (the Bezoutian and the Wronskian of a pencil of binary quadratics); routine.