Find the common factor of \( \Large x^{2}+\frac{1}{x^{2}}=102 \), then the value of \( \Large \left(x-\frac{1}{x}\right) \) is
Correct Answer: Description for Correct answer:
\( \Large x^{2}+\frac{1}{x^{2}} = 102 \)
\( \Large \left(x-\frac{1}{x}\right)^{2} = x^{2}+\frac{1}{x^{2}}-2 \)
= 102 -2 = 100
\( \Large x-\frac{1}{x} = \sqrt{100} = 10 \)
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