Question : If $x+\frac{1}{x}=-14$, and $x<-1$, what will be the value of $x^2-\frac{1}{x^2}$?
Option 1: $-112 \sqrt{3}$
Option 2: $112 \sqrt{3}$
Option 3: $-140 \sqrt{2}$
Option 4: $140 \sqrt{2}$
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Correct Answer: $112 \sqrt{3}$
Solution :
Given, $x+\frac{1}{x}=-14$
Squaring both sides, we get,
⇒ $x^2+\frac{1}{x^2}+2=196$
⇒ $x^2+\frac{1}{x^2}=196-2=194$
Subtracting 2 from both sides, we get,
$x^2+\frac{1}{x^2}-2=194-2$
⇒ $(x-\frac{1}{x})^2=192$
⇒ $x-\frac{1}{x}=-\sqrt{192}$ [as $x<-1$]
Now $x^2-\frac{1}{x^2}$
$=(x-\frac{1}{x})(x+\frac{1}{x})$
$=(-\sqrt{192})\times (-14)$
$=8\sqrt3\times 14 = 112\sqrt3$
Hence, the correct answer is $112\sqrt3$.
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