Question:
The number of elements in the set \(\{x\in\mathbb{R}:(|x|-3)\,|x+4|=6\}\) is equal to :
The number of elements in the set \(\{x\in\mathbb{R}:(|x|-3)\,|x+4|=6\}\) is equal to :
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Split into the cases \(x\ge0\), \(-4\le x<0\) and \(x<-4\), and solve the quadratic in each, keeping only roots that lie in the case.
Updated On: Nov 4, 2026
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The Correct Option is B
Solution and Explanation
Case \(x\ge0\): \((x-3)(x+4)=6\Rightarrow x^2+x-18=0\Rightarrow x=\frac{-1\pm\sqrt{73}}{2}\). Only \(x=\frac{-1+\sqrt{73}}{2}\approx3.77\ge0\) is valid.
Case \(-4\le x<0\): \((-x-3)(x+4)=6\Rightarrow x^2+7x+18=0\), discriminant \(49-72<0\): no real root.
Case \(x<-4\): \((-x-3)(-(x+4))=6\Rightarrow(x+3)(x+4)=6\Rightarrow x^2+7x+6=0\Rightarrow x=-1,-6\). Only \(x=-6\) satisfies \(x<-4\).
Total number of elements \(=2\).
Answer: (2) 2
Case \(-4\le x<0\): \((-x-3)(x+4)=6\Rightarrow x^2+7x+18=0\), discriminant \(49-72<0\): no real root.
Case \(x<-4\): \((-x-3)(-(x+4))=6\Rightarrow(x+3)(x+4)=6\Rightarrow x^2+7x+6=0\Rightarrow x=-1,-6\). Only \(x=-6\) satisfies \(x<-4\).
Total number of elements \(=2\).
Answer: (2) 2
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