The function x^2+ax+2b-a+3 has two distinct real roots for any value a. What is the minimum integral value of b?
+121 The function f(x)=x2+ax+2b−a+3 has distinct real root when the discriminant of the quadratic expression x2+ax+(2b−a+3) is positive. That is when a2−4(2b−a+3)>0; this translates into
a2−4a−12−8b>0. This inequality must hold for any value of a, and so it must hold for the minimum value of the quadratic expression a2−4a−12 , which is equal to - 16 attained at a = - 2 ( think of the vertex of of the parabola defined by the quadratic). So b has to be an integer that results in −16−8b>0. Solving this inequality we get b<−2 and the largest integer smaller than - 2 is - 3.
