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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.

 Jul 30, 2019
 #1
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Since  the interval  [ 1 , 2)   makes this true, then  x = 1  is a root  and  ( x - 1)  must be the linear factor in the numerator

 

Since the function is true for  (-inf, -1)  and [1, 2)

We might guess that we have vertical asymptotes at x  = -1   and  x = 2

So.... (x + 1)  and (x - 2)  seem to be linear factors in the denominator

 

And our function is         ( x - 1)                   ( x  - 1)

                                  ___________  =     ___________

                                   (x + 1) ( x -2)           x^2 - x  - 2

 

See the graph here : https://www.desmos.com/calculator/dhotlrocam

 

Note that  y  is  ≤ 0   on the intervals   (-inf, -1) U  [ 1, 2 )

 

So      a  = -1    b  = -2   and c  = -1

 

And their sum is   - 4

 

 

 

cool cool cool

 Jul 30, 2019
edited by CPhill  Jul 30, 2019

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