Question 622115
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1.  Find the two roots of the analogous quadratic equation and plot the two points.  This quadratic is guaranteed to have two distinct real roots because it is not a perfect square and the signs on the lead coefficient and the constant term are opposite.


2.  Note that you have divided the *[tex \LARGE x]-axis into 3 intervals.


3.  Pick a value from each one of the intervals.  It doesn't matter what the values are so long as they are NOT the endpoints of the interval.


4.  Evaluate the original inequality for each of the three values chosen in step 3.


5.  The interval or intervals that supply the value(s) that make the original inequality true is(are) the interval(s) that comprise the solution set.


6.  Since the original inequality is exclusive of equality (i.e. *[tex \LARGE <] vice *[tex \LARGE \leq]) use parentheses to mark your endpoints when you specify the interval using interval notation.  Your set builder expression must also use *[tex \LARGE <] vice *[tex \LARGE \leq].


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