If a = frac{1}{a-5} and a 0,then the value | vedclass

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(A) Given $a = \frac{1}{a-5}$.Rearranging the equation,we get $a(a-5) = 1$,which simplifies to $a^2 - 5a - 1 = 0$.Dividing the entire equation by $a$

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$\sqrt{29}$

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(A) Given $a = \frac{1}{a-5}$.Rearranging the equation,we get $a(a-5) = 1$,which simplifies to $a^2 - 5a - 1 = 0$.Dividing the entire equation by $a$ (since $a > 0$,$a 
eq 0$),we get $a - 5 - \frac{1}{a} = 0$.This implies $a - \frac{1}{a} = 5$.We need to find $a + \frac{1}{a}$.We know the identity $(a + \frac{1}{a})^2 = (a - \frac{1}{a})^2 + 4$.Substituting the value $a - \frac{1}{a} = 5$,we get $(a + \frac{1}{a})^2 = 5^2 + 4 = 25 + 4 = 29$.Since $a > 0$,$a + \frac{1}{a}$ must be positive.Therefore,$a + \frac{1}{a} = \sqrt{29}$.

If $a = \frac{1}{a-5}$ and $a > 0$,then the value of $a + \frac{1}{a}$ is

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