Problem
ALG-B1-M04-P014 Comparing Two Discriminants
Let \(P\) and \(Q\) be monic quadratic trinomials, each with two distinct roots. The sum of the values of \(Q\) at the roots of \(P\) equals the sum of the values of \(P\) at the roots of \(Q\). Prove that the discriminants of \(P\) and \(Q\) are equal.
Denote the roots of \(P\) by \(a_1,a_2\), and the roots of \(Q\) by \(b_1,b_2\).
Let \(P(x)=(x-a_1)(x-a_2)\), \(Q(x)=(x-b_1)(x-b_2)\). The condition says \(Q(a_1)+Q(a_2)=P(b_1)+P(b_2)\).
Expanding only the needed sums gives \((a_1-b_1)(a_1-b_2)+(a_2-b_1)(a_2-b_2)=(b_1-a_1)(b_1-a_2)+(b_2-a_1)(b_2-a_2)\).
After cancellation this is equivalent to \((a_1-a_2)^2=(b_1-b_2)^2\). For a monic quadratic, the discriminant is the square of the difference of roots. Hence the discriminants are equal.
The hard part is not getting lost in notation; the solution expands only the needed structure.