Problem
ALG-B2-M11-P016 Reversed quadratic
#16
★★★★★ Level 5 of 5
Let \(A x^2+Bx+C>0\) for all real \(x\). Prove that \(C x^2+Bx+A>0\) for all real \(x\).
Hint. Use the sign of the leading coefficient and the discriminant.
The condition gives \(A>0\). Also \(C=P(0)>0\), and the discriminant \(B^2-4AC<0\). For \(C x^2+Bx+A\), the leading coefficient is positive and the discriminant is the same: \(B^2-4CA<0\). Hence it is positive for all real \(x\).
Teaching goal: the student should first recognise the type of estimate, then choose the tool.