If $$\frac{{2p}}{{{p^2} - 2p + 1}} = \frac{1}{4}{\text{,}}$$    p ≠ 0 then the value of $$p + \frac{1}{p}\,{\text{is?}}$$

Solution (By Examveda Team)

$$\eqalign{ & \frac{{2p}}{{{p^2} - 2p + 1}} = \frac{1}{4} \cr & \Rightarrow \frac{{\frac{{2p}}{p}}}{{\frac{{{p^2}}}{p} - \frac{{2p}}{p} + \frac{1}{p}}} = \frac{1}{4} \cr & \Rightarrow \frac{2}{{p + \frac{1}{p} - 2}} = \frac{1}{4} \cr & \Rightarrow p + \frac{1}{p} - 2 = 8 \cr & \Rightarrow p + \frac{1}{p} = 10 \cr} $$