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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