If $\frac{8r}{r^2-8r+1}=\frac{1}{14}$, then the value of $(r +\frac{1}{r})$ is __________.

120

If $\frac{8r}{r^2-8r+1}=\frac{1}{14}$,

then the value of $(r +\frac{1}{r})$ = ?

$\frac{8r}{r^2-8r+1}=\frac{1}{14}$

= 112r = r² - 8r + 1

= r² - 120r + 1

Divide both the sides of the above equation by r then we get,

r + \(\frac{1}{r}\) = 120