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November 28, 2014

November 28, 2014

Posted by **Joe** on Sunday, November 6, 2011 at 11:59am.

- Algebra 3 -
**Steve**, Sunday, November 6, 2011 at 8:07pmIf we put both sides over a common denominator, we have:

(3x-2x^2)/(40x^2) < (3+x)(2+x)/(4-x^2)

collecting terms, we have (3x^2 + 2x + 6)/(4-x^2) > 0

The numerator is always positive, so as long as the denominator is positive, the inequality holds. We stipulated that x<2, and 4-x^2 > 0 if -2<x<2, so that is the interval where the original inequality holds.

- Algebra 3 -
**Steve**, Sunday, November 6, 2011 at 8:08pmsorry. ignore the phrase about stipulating that x<2.

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