What is the value of b+c if $x^2+bx+c>0$ only when $x\in (-\infty, -2)\cup(3,\infty)$?
The quadratic expression x2+bx+c is greater than 0 when its discriminant, b2−4ac, is positive. In this case, b2−4ac>0 only when b2>4ac. This means that b must be positive and c must be negative. Since x2+bx+c>0 for x<−2 and x>3, we know that b<0 and c<0. Therefore, b+c must be negative. The smallest possible value of b+c is −(b+c)=−(−b−c)=b+c=0. Therefore, the value of b+c is 0.