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

The correct option is C $(-7,-4)\cup (-4,1)$

Given inequality is, $\frac{x^2+6x-7}{|x+4|}<0$

Clearly if $x=-4$, then above inequality is not defined

Now other than $-4$, $|x+4|$ is always positive number

So we can multiply above inequality both sides by $|x+4|$

$\Rightarrow x^2+6x-7<0$

$\Rightarrow (x-1)(x+7)<0$

$\Rightarrow -7<x<1$

But $x\ne -4$

Hence required solution is $(-7,-4)\cup (-4,1)$