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 $(-7,-4)\cup (-4,1)$

For $x<-4,$ the inequation becomes
$\frac{x^2+6x-7}{-(x+4)}<0$

$\Rightarrow \frac{(x-1)(x+7)}{(x+4)}>0$

$\Rightarrow x\in (-7,-4)\cup (1,\infty )$
But $x<-4$
$\Rightarrow x\in (-7,-4)$

For $x>-4,$ the inequation becomes
$\frac{x^2+6x-7}{(x+4)}<0$
$\Rightarrow \frac{(x-1)(x+7)}{(x+4)}<0$
$\Rightarrow x\in (-\infty ,-7)\cup (-4,1)$
But $x>-4$
$\Rightarrow x\in (-4,1)$
For $x=-4,$ the expression is undefined.
$\therefore$ The solution set of the inequation in $R$ is $(-7,-4)\cup (-4,1)$