For p, q of degree at most 2, the Wronskian J = p′q − pq′ in coefficients, and J² = d₂p² + d₁pq + d₀q² with d the discriminant of p − Xq

Let $K$ be a commutative ring and $p, q \in K[X]$ of degree at most $2$; write $p_i, q_i$ for their coefficients of $X^i$ and $J = p'q - pq'$ for the Wronskian. Then:

• $J = (p_2q_1 - p_1q_2)X^2 + 2(p_2q_0 - p_0q_2)X + (p_1q_0 - p_0q_1)$;
• $J^2 = d_2p^2 + d_1pq + d_0q^2$ with $d_2 = q_1^2 - 4q_2q_0$, $d_1 = 4(p_2q_0 + p_0q_2) - 2p_1q_1$ and $d_0 = p_1^2 - 4p_2p_0$.

Here $d_2X^2 + d_1X + d_0$ is the discriminant, in $Y$, of $p(Y) - Xq(Y)$. With $[ij] = p_iq_j - p_jq_i$ the identity for $J$ reads $J = [21]X^2 + 2[20]X + [10]$.

Proof. A polynomial of degree at most $2$ is $p_2X^2 + p_1X + p_0$, with $p_i$ its coefficients, so it suffices to prove both identities for two explicit quadratics whose six coefficients are arbitrary elements of $K$. There the derivatives are computed term by term, and both identities are polynomial identities in the six coefficients, closed by ring normalisation. Only the commutativity of $K$ is used.

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