Question : If $2\left [x^{2} +\frac{1}{x^{2}}\right]-2\left [x-\frac{1}{x} \right]-8=0$, what are the two values of $\left (x-\frac{1}{x} \right)\;$?
Option 1: –1 or 2
Option 2: 1 or –2
Option 3: –1 or –2
Option 4: 1 or 2
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Correct Answer: –1 or 2
Solution :
Given: $2\left [x^{2} +\frac{1}{x^{2}} \right]-2\left [ x-\frac{1}{x} \right]-8=0$
$⇒2\left [(x -\frac{1}{x})^{2}+2 \right]-2\left [ x-\frac{1}{x} \right ]-8=0$
Let $(x -\frac{1}{x})=t$
The equation becomes $2(t^{2}+2)-2t-8=0$
$⇒2t^{2}-2t-4=0$
$⇒t^{2}-2t+t-1=0$
$⇒(t-2)(t+1)=0$
$⇒t=-1,2$
So, $(x -\frac{1}{x})=-1,2$
Hence, the correct answer is –1 or 2.
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