Michael J. answered • 01/24/16
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Square both sides of the inequality to undo the square-root.
(2x + 3)(2x + 3) < -x2 - 3x - 2
4x2 + 12x + 9 < -x2 - 3x - 2
Bring all terms to the left side of inequality. This so that we have positive x2 terms.
5x2 + 15x + 11 < 0
Use the quadratic formula to find the zeros:
x = (-b ± √(b2 - 4ac)) / 2a
where:
a = 5
b = 15
c = 11
x = (-15 ± √(225 - 4(55))) / 10
x = (-15 ± √(5)) / 10
x = (-15 ± 2.236) / 10
x = -1.724 and x = -1.276
These are your zeros. Next, we use test points to evaluate the left side of the inequality. Use these test points:
x = -2 , x = -1.5 , and x = 0
The left side of the inequality must be negative. Keep in mind that in the original inequality, the value under the square-root must not negative. Therefore, you will need to do some extensive analysis and evaluations.
I have given you the steps. Now just evaluate using those test point.
Michael J.
From the point where we obtained the zeros, we plug in those values of x into the original inequality. We accept x=-1.724 as a zero because when we plug it back into the original inequality, the inequality is true.
Ignore the test points that I have given you before. Use these new test points instead:
x = -3 , x = -2 , x = -1 , and x = 0
Evaluate the original inequality using these test points. The modified procedure should give you the correct answers, though I am not sure what final answer you put down on your paper.
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01/26/16
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01/25/16