Question 242760
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First thing to do is to solve the problem that you posted before, namely:


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ (x+2)(x-2)(x+1)\ =\ 0]


Take the three roots of this equation, call them *[tex \Large a], *[tex \Large b], and *[tex \Large c] (in ascending numerical order) and create four intervals.


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ \left(-\infty,\,a\right)]


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ \left(a,\,b\right)]


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ \left(b,\,c\right)]


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ \left(c,\,\infty\right)]


For each of the four intervals, select a value and substitute it back into the original inequality.  Once you have done that, two of the values will result in true statements and two of them will make the original inequality false.  The union of the two intervals from which the values that made the original inequality true is the solution set of your inequality.


John
*[tex \LARGE e^{i\pi} + 1 = 0]
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