Question : If $2(x^{2}+\frac{1}{x^{2}})-(x-\frac{1}{x})-7=0$, then two values of $x$ are:
Option 1: $1, 2$
Option 2: $2,-\frac{1}{2}$
Option 3: $0, 1$
Option 4: $\frac{1}{2},1$
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Correct Answer: $2,-\frac{1}{2}$
Solution :
Given: $2(x^{2}+\frac{1}{x^{2}})-(x-\frac{1}{x})-7=0$
We know that, $x^{2}+\frac{1}{x^{2}} = (x-\frac{1}{x})^2+2$
⇒ $2[(x-\frac{1}{x})^2+2]-(x-\frac{1}{x})-7=0$
Put $x-\frac{1}{x}=t$
⇒ $2t^2+4-t-7=0$
⇒ $2t^2-t-3=0$
⇒ $2t^2-3t+2t-3=0$
⇒ $t(2t-3)+1(2t-3)=0$
⇒ $(2t-3)(t+1)=0$
$\therefore t=\frac{3}{2}$ or, $ -1$
Now if,
$x-\frac{1}{x}=\frac{3}{2}$
⇒ $2x^2-2=3x$
⇒ $2x^2-3x-2=0$
⇒ $2x^2-4x+x-2=0$
⇒ $2x(x-2)+1(x-2)=0$
⇒ $(x-2)(2x+1)=0$
$\therefore x = 2$ or, $x = -\frac{1}{2}$
Other two values you can also find by solving $x-\frac{1}{x}=-1$
Hence, the correct answer is $2$ or $-\frac{1}{2}$.
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