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What is the value of b + cif x^2 + bx + c > 0 if only when $x \in (-\infty,2) \cup (3,\infty)$?

 Jun 2, 2021
 #1
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+1

The  function will be  negative   on the interval (2, 3)

 

This  implies  that we  have

 

(x - 2) ( x - 3)  > 0

 

x^2  - 5x  +  6   >   0

 

b =  -5      c  =  6

 

b +  c    =    1

 

 

cool cool cool

 Jun 2, 2021

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