A) \[x\in (-2,8)\]
B) \[x\in (6,16)\]
C) \[x\in (16,\infty )\]
D) \[x\in (-\infty ,-2)\cup (8,\infty )\]
Correct Answer: D
Solution :
(d): Actually, this is a greater than problem, Multiplying both sides by ( ? 1), \[{{x}^{2}}-6x-16>0\Rightarrow (x-8)(x+2)>0\] \[\Rightarrow x\in (-\infty ,-2)\cup (8,\infty )\]
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