If \( \Large x^{2}+\frac{1}{x^{2}}=38, \) find the value of \( \Large x-\frac{1}{x} \)

A) 8
B) 6

C) 4
D) 2

Correct answer:
B) 6

Description for Correct answer:
\( \Large \left(x-\frac{1}{x}\right)^{2} = x^{2}+\frac{1}{x^{2}}-2 \)

= 38 - 2 = 36

Therefore, \( \Large x-\frac{1}{x} = \sqrt{36} = 6 \)

Please provide the error details in above question

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