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Mathematics Question on Complex Numbers and Quadratic Equations

The solution set of the inequation x2+6x−7∣x+4∣<0\frac {x^2+6x-7}{|x+4|}<0 is

A

(−7,1)(-7,1)

B

(−7,−4)(-7,-4)

C

(−7,−4)∪(−4,1)(-7, - 4) \cup (- 4, 1)

D

(−7,−4)∪(4,1)(-7, - 4) \cup (4, 1)

Answer

(−7,−4)∪(−4,1)(-7, - 4) \cup (- 4, 1)

Explanation

Solution

x2+6x−7∣x+4∣<0\frac{x^{2}+6x-7}{\left|x+4\right|}<0
We know that ∣x+4∣> 0|x + 4| >\, 0
⇒x2+6x−7< 0\Rightarrow x^{2} + 6x - 7 < \,0 and x≠−4x \ne- 4
⇒(x+7)(x−1)< 0\Rightarrow \left(x + 7\right)\left(x - 1\right) < \,0 and x≠−4x \ne- 4
\Rightarrow x\,\in\,\left(-7,1\right)-\left\\{-4\right\\}
⇒x ∈ (−7,−4)∪(−4,1)\Rightarrow x\,\in\,\left(-7,-4\right)\cup\left(-4, 1\right)