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

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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
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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