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

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

Mar 2, 2018
edited by Melody  Mar 2, 2018

#1
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What is the value of   $$b+c$$   if     $$x^2+bx+c>0$$     only when    $$x\in (-\infty, -2)\cup(3,\infty)$$?

Consider the concave up parabola   y=x^2+bx+c

$$x^2+bx+c>0$$    will be true when y is positive, which is when the graph is above the x axis.

The roots of this graph are

$$x=\frac{-b\pm\sqrt{b^2-4c}}{2}\\ x=\frac{-b}{2}\pm\frac{\sqrt{b^2-4c}}{2}\\$$

the axis of symmetry of this graph will be     $$x=\frac{-b}{2}$$

No consider the domain where x is poistive.

It is less than -2 and greater then 3  SO the axis of symmetry must be half way between these points.  (-2+3)/2=-1/2

so

$$\frac{-b}{2}=\frac{-1}{2}\\ b=1$$

Also

$$\frac{-b}{2}+\frac{\sqrt{b^2-4c}}{2}=3\\ -b+\sqrt{b^2-4c}=6\\ sub\;in\;b=1\\ -1+\sqrt{1^2-4c}=6\\ \sqrt{1-4c}=7\\ 1-4c=49\\ -4c=48\\ c=-12\\ ~\\ b+c=1+-12=-11$$

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Mar 2, 2018
#2
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Sorry Melody, but the axis of symmetry is x = 1/2 not -1/2.

Also, it's quicker to say that the equation of the parabola must be

$$y=(x+2)(x-3)=x^2-x-6$$

so b + c =  -7.

Mar 2, 2018