Question

The solution set of the inequation $\frac{x^2+6x-7}{|x+4|}<0$ is

• A. $(-7,1)$
• B. $(-7,-4)$
• C. $(-7,-4)\cup (-4,1)$
• D. $(-7,-4)\cup (4,1)$

Answer 1

Answer: (C)

$\frac{x^2+6x-7}{|x+4|}<0$
We know that $|x+4|>0$
$\Rightarrow x^2+6x-7<0$ and $x\ne -4$
$\Rightarrow \left(x+7\right)\left(x-1\right)<0$ and $x\ne -4$
$\Rightarrow x\in \left(-7,1\right)-\left\{-4\right\}$
$\Rightarrow x\in \left(-7,-4\right)\cup \left(-4,1\right)$