Question : If $x^4+\frac{16}{x^4}=15617, x>0$, then find the value of $x+\frac{2}{x}$.
Option 1: $\sqrt{121}$
Option 2: $\sqrt{129}$
Option 3: $\sqrt{123}$
Option 4: $\sqrt{127}$
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Correct Answer: $\sqrt{129}$
Solution : Given: $x^4+\frac{16}{x^4}=15617$ ⇒ $(x^2)^2 + (\frac{4}{x^2})^2 + 2\times x^2 \times \frac{4}{x^2} = 15617 + 2\times x^2 \times \frac{4}{x^2}$ ⇒ $(x^2+\frac{4}{x^2})^2 = 15625$ ⇒ $x^2 + \frac{4}{x^2} = 125$ ⇒ $x^2 + (\frac{2}{x})^2 + 2\times x \times \frac{2}{x} = 125 + 2\times x\times \frac{2}{x}$ ⇒ $(x+\frac{2}{x})^2 = 129$ ⇒ $x+\frac{2}{x} = \sqrt{129}$ Hence, the correct answer is $\sqrt{129}$.
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