Question:

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

Updated On: Apr 15, 2024
  • (−7,1)(-7,1)
  • (−7,−4)(-7,-4)
  • (−7,−4)∪(−4,1)(-7, - 4) \cup (- 4, 1)
  • (−7,−4)∪(4,1)(-7, - 4) \cup (4, 1)
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The Correct Option is C

Solution and Explanation

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
⇒x ∈ (−7,1)−{−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)
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Concepts Used:

Complex Numbers and Quadratic Equations

Complex Number: Any number that is formed as a+ib is called a complex number. For example: 9+3i,7+8i are complex numbers. Here i = -1. With this we can say that i² = 1. So, for every equation which does not have a real solution we can use i = -1.

Quadratic equation: A polynomial that has two roots or is of the degree 2 is called a quadratic equation. The general form of a quadratic equation is y=ax²+bx+c. Here a≠0, b and c are the real numbers.