Question

(a) Find the values of x for which the following inequality holds $\frac{8x^2+16x-51}{(2x-3)(x+4)}>3$.
(b) Find the values of x, which satisfy the inequality
$\frac{x-2}{x+2}>\frac{2x-3}{4x-1}$.

Answer 1

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

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 inquality as under :

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

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

Writing the above as under

or $\frac{2[x-(-3\left)\right](x-5/2)}{2[x-(-4\left)\right](x-3/2)}>0$

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

$\therefore$ $N^r$ is $>0$ 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.

From the above fig. it is clear that $N^r$ is $+ive$ for

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

(b) $\frac{x-2}{x+2}-\frac{2x-3}{4x-1}>0$

or $\frac{2(x^2-5x+4)}{(x+2)(4x-1)}>0$

or $\frac{2(x-4)(x-1)}{(x+2)(4x-1)}>0$

or $\frac{2(x-1)(x-4)}{4[x-(-2\left)\right](x-\frac{1}{4})}>0$

Now proceed as in part (a). Then

$x<-2$ or $\frac{1}{4}<x$ or $x>4$.