Question

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

Answer 1

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

Here we cannot write $8x^2+16x-51>3(2x-3)(x+4)$ as in inequalities we can multiply both sides only by + ive quantity. But here we do not know whether $(2x-3)(x+4)$ is +ive or-ive.

Hence we write the inequality as under:

$\frac{8x^2+16x-51}{2x^2+5x-12}-3>0$

simplifying

$\frac{2x^2+x-15}{2x^{2}5x-12}>0$ or $\frac{(2x-5)(x+3)}{(2x-3)(x+4)}>0$

Writing the above as under or $\frac{2[x-\left(3\right)](x-5/2)}{2(x-(-4)(x-3/2))}>0$ or $\frac{(2x-5)(x+3)(x+4)}{{(2x-3)}^2{(x+4)}^2}>0$

$\therefore N^{r}is0>as{D}^{r}is+ive.$ The values of x obtained from $N^r=0$ are $\frac{5}{2},-3,\frac{3}{2},-4$ Mark them in ascending order on real line as shown below. Write + in the extreme right and move towards left with opposite signs in successive intervals

.Form the above fig. it is clear that $N^r$ is +ive for $x>\frac{5}{2},-3<x<\frac{3}{2},x<-4$