+118609 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
\(\frac{-b}{2}+\frac{\sqrt{b^2-4c}}{2}=3\\ -b+\sqrt{b^2-4c}=6\\ sub\;in\;b=1\\ -1+\sqrt{1^2-4c}=6\\ \sqrt{1-4c}=7\\ 1-4c=49\\ -4c=48\\ c=-12\\ ~\\ b+c=1+-12=-11\)
