What is the value of b + c if $x^2+bx+c>0$ only when $x\in (-\infty, -2)\cup(3,\infty)$?
The quadratic x2+bx+c is positive when its discriminant, b2−4ac, is positive. So we have b2−4ac>0. Expanding the discriminant, we get b2−4ac=(b+2)(b−3)>0. This means that b+2>0 and b−3<0, or b>−2 and b<3. The only value of b that satisfies both of these inequalities is b=2.
Substituting b=2 into the discriminant, we get 4−4c>0, so c<1. The smallest possible value of c is 0, so b+c=2.
To check our answer, we can plug in b=2 and c=0 into the quadratic and see that it is positive when x<−2 and x>3.
