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?

Guest Dec 16, 2019

#1**0 **

The function \(f(x)=x^2+ax+2b-a+3\) has distinct real root when the discriminant of the quadratic expression \(x^2+ax+(2b-a+3)\) is positive. That is when \(a^2-4(2b-a+3)>0\); this translates into

\(a^2-4a-12-8b>0\). This inequality must hold for any value of a, and so it must hold for the minimum value of the quadratic expression \(a^2-4a-12\) , 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.

Gadfly Dec 17, 2019