If frac{{2x}}{{2{x^2} + 5x + 2}} > frac{1}{{x + 1}} , then

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4-2.Quadratic Equations and Inequations

medium

If$\frac{{2x}}{{2{x^2} + 5x + 2}} > \frac{1}{{x + 1}}$, then

$ - 2 > x > - 1$

$ - 2 \ge x \ge - 1$

$ - 2 < x < - 1$

$ - 2 < x \le - 1$

(IIT-1987)

Solution

==> $\frac{{2x}}{{(2x + 1)(x + 2)}} > \frac{1}{{(x + 1)}}$

==> $\frac{{2x}}{{(2x + 1)(x + 2)}} – \frac{1}{{(x + 1)}} > 0$

==> $\frac{{2x(x + 1) – (2x + 1)(x + 2)}}{{(x + 1)(2x + 1)(x + 2)}} > 0$

==> $\frac{{2{x^2} + 2x – 2{x^2} – 4x – x – 2}}{{(x + 1)(x + 2)(2x + 1)}} > 0$

==> $\frac{{ – 3x – 2}}{{(x + 1)(x + 2)(2x + 1)}} > 0$

Equating each factor equal to $0,$ we have $x = – 2, – 1, – \frac{2}{3}, – \frac{1}{2}$.
It is clear that $ – \frac{2}{3} < x < – \frac{1}{2}$ or $ – 2 < x < – 1$.

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