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

 Sep 10, 2022
 #1
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\( x\in (-\infty, -2)\cup(3,\infty)\)

 

So we have \((-2)^2 -2b + c = 0\), meaning \(-2b + c = -4\)

 

We also have \((3)^2 + 3b + c = 0\), meaning \(3b + c = -9\)

 

Solving the system gives us \(b = -1\) and \(c = -6\).

 

Thus, \(b + c = -1 - 6 = \color{brown}\boxed{-7}\)

 Sep 11, 2022
 #2
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thank you!

 Sep 11, 2022

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