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\sqrt{2x^2 +ax+b} > x-c <=> x \in (-\infty,0)\cup(1,\infty)

Source: Czech And Slovak Mathematical Olympiad, Round III, Category A 2001 p3

February 11, 2020
inequalitiesradicalalgebra

Problem Statement

Find all triples of real numbers (a,b,c)(a,b,c) for which the set of solutions xx of 2x2+ax+b>x−c\sqrt{2x^2 +ax+b} > x-c is the set (−∞,0]∪(1,∞)(-\infty,0]\cup(1,\infty).
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