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

 Jun 21, 2020
 #1
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b + c = 4 + (-7) = -3.

 Jun 22, 2020
 #2
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That means \(x^2 + bx + c \leqslant 0\) when \(x\in \left[-2, 3\right]\).

 

And this also means \(x^2 + bx + c < 0\) when \(x\in \left(-2, 3\right)\).

 

When \(x^2 + bx + c = 0\)\(x = -2\text{ or }x = 3 \)

 

So \(x^2 + bx + c = (x + 2)(x - 3) = x^2 - x - 6\).

 

Comparing coefficients, b = -1 and c = -6.

 

b + c = -1 + (-6) = -7.

 Jun 22, 2020

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