Question : If $x^{4}+\frac{1}{x^{4}}=119$, then the value of $x-\frac{1}{x}$ is:
Option 1: 6
Option 2: 12
Option 3: 11
Option 4: 3
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Correct Answer: 3
Solution :
Given: $x^4+\frac{1}{x^4} = 119$
⇒ $(x^2+\frac{1}{x^2})^2–2×x^2×\frac{1}{x^2} = 119$
⇒ $(x^2+\frac{1}{x^2})^2–2 = 119$
⇒ $(x^2+\frac{1}{x^2})^2 = 119+2 = 121$
⇒ $x^2+\frac{1}{x^2} = \sqrt{121} = 11$
Now, $(x–\frac{1}{x})^2+2×x×\frac{1}{x} = 11$
⇒ $(x–\frac{1}{x})^2+2 = 11$
⇒ $(x–\frac{1}{x})^2 = 11–2 = 9$
⇒ $x–\frac{1}{x} = \sqrt9 = 3$
Hence, the correct answer is 3.
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