Find the value of 'p' for which one root of the quadratic equation $\displaystyle \mathrm{px}^{2}-14 \mathrm{x}+8=0$ is $\displaystyle 6$ times the other.

Let roots of the quadratic equation be \(\displaystyle \alpha, 6 \alpha\) \[\begin{aligned} & \mathrm{px}^{2}-14 \mathrm{x}+8=0 \\ & \therefore \alpha+6 \alpha=\frac{14}{\mathrm{p}} \Rightarrow 7 \alpha=\frac{14}{\mathrm{p}} \Rightarrow \alpha=\frac{2}{\mathrm{p}} \\ & \text { and } \alpha \cdot 6 \alpha=\frac{8}{\mathrm{p}} \Rightarrow 6 \alpha^{2}=\frac{8}{\mathrm{p}} \Rightarrow 6 \cdot \frac{4}{\mathrm{p}^{2}}=\frac{8}{\mathrm{p}} \\ & \Rightarrow \mathrm{p}=3 \end{aligned} \]