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$x^4-x^2+1=0$ হলে, $x^2+\frac{1}{x^2}$ এর মান কত?

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দেওয়া আছে,
$x^4-x^2+1=0$

বা, $\frac{x^4-x^2+1}{x^2}=\frac{0}{x^2}$
[উভয় পার্শ্বে $x^2$ দ্বারা ভাগ করে]

বা, $\frac{x^4}{x^2}-\frac{x^2}{x^2}+\frac{1}{x^2}=0$

বা, $x^2-1+\frac{1}{x^2}=0$

$\therefore x^2+\frac{1}{x^2}=1$ [Answer]

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