Question : If $x^2-7 x+1=0$ and $0<x<1$, what is the value of $x^2-\frac{1}{x^2}?$
Option 1: $21\sqrt{5}$
Option 2: $-21\sqrt{5}$
Option 3: $28\sqrt{5}$
Option 4: $-28\sqrt{5}$
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Correct Answer: $-21\sqrt{5}$
Solution :
Given: $x^2-7 x+1=0$
⇒ $x+\frac{1}{x}=7$
Squaring the equation, we get,
$(x+\frac{1}{x})^2=7^2$
⇒ $x^2+\frac{1}{x^2}+2=49$
⇒ $x^2+\frac{1}{x^2}=47$
And we know, $(x-\frac{1}{x})^2=x^2+\frac{1}{x^2}-2$
⇒ $(x-\frac{1}{x})^2=47-2=45$
⇒ $x-\frac{1}{x}=\pm\sqrt{45}$
⇒ $x=-\sqrt{45}$ [As $0<x<1$]
Now consider, $x^2-\frac{1}{x^2}$
Using $a^2-b^2=(a-b)(a+b)$
$=(x+\frac{1}{x})(x-\frac{1}{x})$
$=7\times (-\sqrt{45})$
$=-7\times 3\sqrt5=-21\sqrt5$
Hence, the correct answer is $-21\sqrt5$.
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