Question : If $\left(x^2+\frac{1}{x^2}\right)=6$ and $0<x<1$, what is the value of $x^4-\frac{1}{x^4}$?
Option 1: $24\sqrt{2}$
Option 2: $-24\sqrt{2}$
Option 3: $-12\sqrt{10}$
Option 4: $12\sqrt{10}$
Correct Answer: $-24\sqrt{2}$
Solution :
Use (a – b)
2
= (a + b)
2
– 4ab
So, $(x^2-\frac{1}{x^2})^2 = (x^2+\frac{1}{x^2})^2 -4$
⇒ $(x^2-\frac{1}{x^2})^2 = 6^2 -4$
⇒ $(x^2-\frac{1}{x^2}) = - \sqrt {32}\ \ \ [\because 0<x<1]$
Also, $x^4 - \frac1{x^4} = (x^2+\frac{1}{x^2})(x^2-\frac{1}{x^2})$
⇒ $x^4 - \frac1{x^4} = - 6 \times \sqrt{32}$
$\therefore x^4 - \frac1{x^4} = - 24 \sqrt{2}$
Hence, the correct answer is $-24\sqrt2$.
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