Question

The set of all real values of $x$ for which $\frac{8x^2+16x-51}{(2x-3)(x+4)}<3$ is

• A. $(-3,5/2)$
• B. $(-4,-3)$
• C. Both A and B
• D. None of these

Answer 1

Answer: (A)

$(-3,5/2)$

Given,

$\frac{8x^2+16x-51}{(2x-3)(x+4)}<3$

or, $8x^2+16x-51<3(2x^2+5x-12)$

or, $2x^2+x-15<0$

or, $2x^2+6x-5x-15<0$

or, $2x(x+3)-5(x+3)<0$

or, $(2x-5)(x+3)<0$

or, $x\in (-3,\frac{5}{2})$.