The values of x for which the inequality 8x2+16x-512x-3x+4>3 holds, are

Given 8x2+16x−51(2x−3)(x+4)>3⇒ 8x2+16x−51(2x−3)(x+4)−3>0⇒ 8x2+16x−51−6x2−15x+36(2x−3)(x+4)>0⇒ 2x2+x−15(2x−3)(x+4)>0⇒ 2x2+6x−5x−15(2x−3)(x+4)>0⇒ (x+3)(2x−5)(2x−5)(x+4)>0∴ (2x−3)(x+4)(x+3)(2x−5)>0∴ x∈ (−∞, −4)∪(−3, 32)∪(52, ∞)

The values of x for which the inequality 8x2+16x-512x-3x+4>3 holds, are

Given 8x2+16x−51(2x−3)(x+4)>3⇒ 8x2+16x−51(2x−3)(x+4)−3>0⇒ 8x2+16x−51−6x2−15x+36(2x−3)(x+4)>0⇒ 2x2+x−15(2x−3)(x+4)>0⇒ 2x2+6x−5x−15(2x−3)(x+4)>0⇒ (x+3)(2x−5)(2x−5)(x+4)>0∴ (2x−3)(x+4)(x+3)(2x−5)>0∴ x∈ (−∞, −4)∪(−3, 32)∪(52, ∞)

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The values of x for which the inequality $\frac{8 x^{2} + 16 x - 51}{(2 x - 3) (x + 4)} > 3$ holds, are

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$x \geq 4$

$\frac{3}{2} < x < \frac{5}{2} x < 4 or x < \frac{5}{2} or - 3 < x < \frac{3}{2}$

$x < - 4 or x > \frac{5}{2} or - 3 < x < \frac{3}{2}$

$- 4 \leq x \leq - 3$

answer is D.

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Detailed Solution

$\begin{array}{l} G i v e n \frac{8 x^{2} + 16 x - 51}{(2 x - 3) (x + 4)} > 3 \\ \Rightarrow \frac{8 x^{2} + 16 x - 51}{(2 x - 3) (x + 4)} - 3 > 0 \\ \Rightarrow \frac{8 x^{2} + 16 x - 51 - 6 x^{2} - 15 x + 36}{(2 x - 3) (x + 4)} > 0 \\ \Rightarrow \frac{2 x^{2} + x - 15}{(2 x - 3) (x + 4)} > 0 \\ \Rightarrow \frac{2 x^{2} + 6 x - 5 x - 15}{(2 x - 3) (x + 4)} > 0 \\ \Rightarrow \frac{(x + 3) (2 x - 5)}{(2 x - 5) (x + 4)} > 0 \\ ∴ (2 x - 3) (x + 4) (x + 3) (2 x - 5) > 0 \\ ∴ x \in (- \infty, - 4) \cup (- 3, \frac{3}{2}) \cup (\frac{5}{2}, \infty) \end{array}$