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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 \(a+b+c\)

 

Please help! Answer is NOT -1

 Apr 13, 2020
 #1
avatar+21004 
+1

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

 Apr 13, 2020
 #2
avatar+25 
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isnt the question asking for what a+b+c is?

caadfo1118  Apr 13, 2020
 #3
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i'm looking for a value though.....

Guest Apr 13, 2020

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