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# much help appreciated!

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The solution to the inequality $$\frac{x + c}{x^2 + ax + b} \le 0$$  is $$x \in (-\infty,-1) \cup [1,2).$$ Find $$a+b+c$$

Apr 13, 2020

#1
+21004
+1

Try this:  (x - 1) / [ (x + 1)·(x - 2) ]  <= 0

Apr 13, 2020
#2
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isnt the question asking for what a+b+c is?