Asked by Joe
Trying to solve this rational inequality but can't figure how to factor the numerator. ((3x-2x^2)/(4-x^2))<((3+x)/(2-x))
Answers
Answered by
Steve
If 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.
(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.
Answered by
Steve
sorry. ignore the phrase about stipulating that x<2.
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