Question 969305
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*[tex \LARGE \ \ \ \ \ \ \ \ \ \  (x\,+\,3)(x\,-\,4)\ <\ 7]


*[tex \LARGE \ \ \ \ \ \ \ \ \ \  x^2\ -\ x\ -\ 12\ <\ 7]


*[tex \LARGE \ \ \ \ \ \ \ \ \ \  x^2\ -\ x\ -\ 19\ <\ 0]


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ x\ =\ \frac{-(-1)\ \pm\ \sqrt{(-1)^2\ -\ 4(1)(-19)}}{2(1)}\ =\ \frac{1\ \pm\ \sqrt{77}}{2}]


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ \frac{1\ -\ \sqrt{77}}{2}\ \approx\ -3.9]


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ \frac{1\ +\ \sqrt{77}}{2}\ \approx\ 4.9]


Divide the *[tex \Large x]-axis into three regions:


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ \left(-\infty,\frac{1\ -\ \sqrt{77}}{2}\right)]


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ \left(\frac{1\ -\ \sqrt{77}}{2},\frac{1\ +\ \sqrt{77}}{2}\right)]


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ \left(\frac{1\ -\ \sqrt{77}}{2},\infty\right)]


Pick a value from each interval:  -4, 0, and 5 will do nicely.


Test each value in the original inequality:


*[tex \LARGE \ \ \ \ \ \ \ \ \ \  (-4\,+\,3)(-4\,-\,4)\ =\ 8\ <\ 7]


Is a False statement.


*[tex \LARGE \ \ \ \ \ \ \ \ \ \  (0\,+\,3)(0\,-\,4)\ =\ -12 <\ 7]


Is a True statement.


*[tex \LARGE \ \ \ \ \ \ \ \ \ \  (5\,+\,3)(5\,-\,4)\ =\ 8\ <\ 7]


Is a False statement.


Hence, the inequality is true for values of *[tex \Large x] in the interval


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ \left(\frac{1\ -\ \sqrt{77}}{2},\frac{1\ +\ \sqrt{77}}{2}\right)]


Therefore, the solution set of the inequality is


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ \left{x\,\in\,\mathbb{R}\,:\,\frac{1\ -\ \sqrt{77}}{2}\ <\ x\ <\ \frac{1\ +\ \sqrt{77}}{2}\right}]


John
*[tex \LARGE e^{i\pi}\ +\ 1\ =\ 0]
My calculator said it, I believe it, that settles it

*[tex \Large \ \
*[tex \LARGE \ \ \ \ \ \ \ \ \ \