If $\frac{8r}{r^2-8r+1}=\frac{1}{14}$, then the value of $(r +\frac{1}{r})$ is __________.
Answer & explanation
Correct answer: option 3
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