Question

The values of $x$, which satisfies $\frac{8x^2+16x-51}{(2x-3)(x+4)}>3$, is

• A. $x<-4$
• B. $-3<x<\frac{3}{2}$
• C. $x>\frac{5}{2}$
• D. All the above

Answer 1

The correct option is D All the above

Given, $\frac{8x+16x-51}{2x-3\left)\right(x+4}>3$
$\Rightarrow \frac{8x+16x-51}{2x+5x-12}>3$
$\Rightarrow \frac{8x+16x-51}{2x+5x-12}-3>0$
$\Rightarrow \frac{(8x+16x-51)-(6x-15x+36)}{(2x-3)(x+4)}>0$
On simplification, we get
$\frac{2x+x-15}{(2x-3)(x+4)}>0$
$\Rightarrow \frac{(2x-5)(x+3)}{(2x-3)(x+4)}>0$ ...(i)
Therefore from equation (i).
The answer is all of the above (D).