Problem
ALG-B2-M01-P020 Roots and an interval
Let \(b>0\), and suppose the quadratic trinomial \(x^2+ax+b\) has two distinct real roots. Exactly one root lies in the segment \([-1,1]\). Prove that exactly one root lies in the interval \((-b,b)\).
Hint 1. Denote the roots by \(r,s\).
Hint 2. Use \(rs=b>0\), so the roots have the same sign.
Let the roots be \(r,s\), and suppose \(|r|\le1\), \(|s|>1\), since exactly one root lies in \([-1,1]\). By Vieta, \(rs=b>0\), so the roots have the same sign and \(b=|r||s|\). Since \(|s|>1\), we get \(|r|
A. Source analysis. Main objects: inequalities, order, an extremal element, or an invariant. The obvious first move usually gives only a local estimate. The hidden observation is to choose the right nondecreasing quantity, or to add/multiply inequalities only after signs are controlled. The needed step is an ordering, an invariant, a product transformation, or a boundary case.
F. Difficulty justification. Final level 8: one must connect root location with their product via Vieta.
G. Why this is not a one-step exercise. The sign of the trinomial at the endpoints does not directly give the interval \((-b,b)\).