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Comparison of Quantities Using Exponents

Solve x2 + ...

Question

Solve
$(x^{2} + 3 x) (2 x + 3) - 16 \frac{2 x + 3}{x^{2} + 3 x} \geq 0$

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Solution

$x^{2} + 3 x + 4 = x^{2} + 2 (x) \frac{3}{2} + \frac{9}{4} + \frac{7}{4} = (x + \frac{3}{2})^{2} + \frac{7}{4} > 0$
$(x^{2} + 3 x) (2 x + 3) - 16 \frac{2 x + 3}{x^{2} + 3 x} \geq 0$
$\frac{2 x + 3}{x^{2} + 3 x} ((x^{2} + 3 x)^{2} - 4^{2}) \geq 0$
$\frac{(2 x + 3) (x^{2} + 3 x)}{(x^{2} + 3 x)^{2}} (x^{2} + 3 x + 4) (x^{2} + 3 x - 4) \geq 0$
$x (x + 3) (2 x + 3) (x - 1) (x + 4) \geq 0$
$⟹ x \in [- 4, - 3] \cup [- \frac{3}{2}, 0] \cup [1, \infty)$

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