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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 \(a + b + c. \)

 

aHHH I'm so stuckkkk

 Mar 30, 2020
 #1
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a + b + c = (-3) + 4 + (-2) = -1.

 Mar 31, 2020
 #2
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it's wrong for some reason....i have another attempt...help please!

Guest Apr 1, 2020

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