Question

Solve the inequality :

$\frac{(2x^2+3)(x+2)}{(x^2-4)}\ge 0$

• A. no solutions
• B. $(-\infty ,2)$
• C. $(2,+\infty )$
• D. $[2,+\infty )$

Answer 1

The correct option is C $(2,+\infty )$

Given $\frac{(2x^2+3)(x+2)}{(x^2-4)}\ge 0$

In the fraction $2x^2+3$ is always greater than 0

Remaining fraction

$=\frac{x+2}{x^2-4}$

$=\frac{x+2}{(x-2)(x+2)}$

$\Rightarrow \frac{1}{x-2}\ge 0$ subjected to the condition $x\ne 2$

$\Rightarrow x\ge 2$ but $x\ne 2$

$\therefore x>2$

$\Rightarrow x\in (2,\infty )$