Problem
ALG-B2-M10-P018 Two quadratic trinomials
Let \(A,B,C\) be real numbers, and suppose \[A x^2+(B-C)x+C>0\] for all real \(x\). Prove that \[C x^2-(B+C)x+(A+B)>0\] for all real \(x\).
Hint 1. The first trinomial has negative discriminant.
Hint 2. Compute the discriminant of the second trinomial.
From the condition at \(x=0\), we get \(C>0\). The first trinomial is positive for all \(x\), so its discriminant is negative: \[(B-C)^2-4AC<0.\] The discriminant of the second trinomial is \[(B+C)^2-4C(A+B)=B^2-2BC+C^2-4AC=(B-C)^2-4AC<0.\] Its leading coefficient is \(C>0\), hence the second trinomial is also positive for all \(x\).
A. Source analysis. Main objects: two quadratic trinomials connected by a coefficient transformation.
B. Insufficient first move. Testing separate values of \(x\) cannot prove positivity for all real \(x\).
C. Hidden observation. The second trinomial has the same discriminant as the first.
D. Required move. One must translate positivity into the sign of the discriminant.
E. Number of ideas. Two ideas: discriminant and positivity of the leading coefficient.
F. Difficulty justification. Regional level 7: the solution requires seeing the invariant discriminant, not pointwise checking.
G. Why it is not one-step. It is not one-step: one must extract \(C>0\) and \(D<0\), then transfer them to the second trinomial.