Question 1031398
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First, solve the analogous equation:


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ x^2\ +\ 6x\ +\ 8\ =\ 0]


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ (x\ +\ 2)(x\ +\ 4)\ =\ 0]


So *[tex \Large x\ =\ -2\ ] or *[tex \Large x\ =\ -4]


Note that the two roots of the equation divide the *[tex \Large x]-axis into three regions:  *[tex \Large (-\infty,-4),\ ]*[tex \Large (-4,-2),\ ] and *[tex \Large (-2,\infty)]


Choose a value from the interior of each of the three intervals, -5, -3, and 0 should do nicely.


Evaluate *[tex \Large x^2\ +\ 6x] for each of the three selected values.  One result is less than -8.  The interval from which this value was taken is the solution set of the original inequality.  Remember to make both endpoints inclusive because of the *[tex \Large \leq] in the original inequality.


Use a graphing calculator to check the validity of your result.  If you graph *[tex \Large x^2\ +\ 6x\ +\ 8] a portion of the curve should be below the x-axis.  The endpoints of this portion of the curve are the endpoints of your solution set interval. 


*[illustration x_2_6x_8_and_leq_zero_solution.jpg]


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

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

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