Question : If $\left(x-\frac{1}{x}\right)^2=12$, what is the value of $\left(x^2-\frac{1}{x^2}\right)$, given that $x>0$?
Option 1: $6 \sqrt{2}$
Option 2: $8 \sqrt{3}$
Option 3: $6 \sqrt{3}$
Option 4: $8 \sqrt{2}$
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Correct Answer: $8 \sqrt{3}$
Solution :
Given,
$\left(x-\frac{1}{x}\right)^2=12$
⇒ $x^2+\frac{1}{x^2}=12+2$
⇒ $x^2+\frac{1}{x^2}+2=12+2+2$
⇒ $x^2+\frac{1}{x^2}+2(x)(\frac{1}{x})=12+2+2$
⇒ $(x+\frac{1}{x})^2=16$
⇒ $(x+\frac{1}{x})=4$
And, $\left(x-\frac{1}{x}\right)=2\sqrt3$
Now,$\left(x^2-\frac{1}{x^2}\right)$
= $(x-\frac{1}{x})(x+\frac{1}{x})$
= $4\times2\sqrt3$
= $8\sqrt3$
Hence, the correct answer is $8\sqrt3$.
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