Question : If $x=\frac{1}{x-5}(x>0)$, then the value of $x+\frac{1}{x}$ is:
Option 1: $\sqrt{41}$
Option 2: $\sqrt{29}$
Option 3: $\sqrt{23}$
Option 4: $\sqrt{43}$
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Correct Answer: $\sqrt{29}$
Solution : $x=\frac{1}{x-5}(x>0)$ $⇒x^2-5x-1=0$ $⇒x^2-1=5x$ Multiplying both sides by $\frac{1}{x}$, we get, $⇒x-\frac{1}{x}=5$ Squaring both sides, we get, $⇒(x-\frac{1}{x})^2=5^2$ $⇒x^2+\frac{1}{x^2}-2=25$ Adding 4 to both sides, we get, $⇒x^2+\frac{1}{x^2}+2=25+4$ $⇒(x+\frac{1}{x})^2=29$ $\therefore x+\frac{1}{x}=\sqrt{29}$ Hence, the correct answer is $\sqrt{29}$.
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