The solution to the inequality \(\frac{x + c}{x^2 + ax + b} \le 0 \) is \(x \in (-\infty,-1) \cup [1,2). \) Find \(a+b+c\)
Please help! Answer is NOT -1
Try this: (x - 1) / [ (x + 1)·(x - 2) ] <= 0
isnt the question asking for what a+b+c is?
i'm looking for a value though.....
