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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

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