Question

Find the sum of all the real solutions of the inequality $\frac{\left(x^2+2\right)(\sqrt{x^2-16})}{\left(x^4+2\right)\left(x^2-9\right)}\le 0$ .

Answer 1

Answer: 0

Given,
$\frac{\left(x^2+2\right)(\sqrt{x^2-16})}{\left(x^4+2\right)\left(x^2-9\right)}\le 0$
Here, $\left(x^2+2\right)$ and $\left(x^4+2\right)$ are positive $\forall x\in R$ .
Also, $\sqrt{x^2-16}\ge 0$
Then, $\left(x^2-9\right)<0$
$\Rightarrow \left(x-3\right)\left(x+3\right)<0$
$\Rightarrow -3<x<\mathrm{3...}(1)$

Now, for $\sqrt{x^2-16}$ to be defined,
$x^2-16\ge 0$
$\Rightarrow x\le -4$ or $x\ge \mathrm{4...}(2)$
From $(1)$ and $(2)$ ,
$x=\varphi$
But for equality in the given problem, numerator can be zero,
i.e., $x^2-16=0$
$\Rightarrow x=\pm 4$
Hence, sum of all the real solution is $4-4=0$ .