Find the value of 'p' for which one root of the quadratic equation px2 - 14x + 8 = 0 is 6 times the other.

Let roots of the quadratic equation be $$\begin{aligned}& \alpha, 6\alpha \\ & px^2 - 14x + 8 = 0 \\ & therefore \alpha + 6\alpha = -\frac{-14}{p} \Rightarrow 7\alpha = \frac{14}{p} \Rightarrow \alpha = \frac{2}{p}\end{aligned}$$and $$\begin{aligned}& \alpha \cdot 6\alpha = \frac{8}{p} \Rightarrow 6\alpha^2 = \frac{8}{p} \Rightarrow 6 \cdot (\frac{2}{p})^2 = \frac{8}{p} \\ & \Rightarrow 6 \cdot \frac{4}{p^2} = \frac{8}{p} \\ & \Rightarrow \frac{24}{p^2} = \frac{8}{p} \\ & \Rightarrow 24p = 8p^2 \Rightarrow 8p^2 - 24p = 0 \Rightarrow 8p(p-3) = 0 \\ & \Rightarrow p = 0 \text{ or } p = 3 \\ & \text{Since } p \neq 0 \text{ (coefficient of } x^2 \text{ cannot be zero)} \\ & \Rightarrow p = 3\end{aligned}$$