Find the value of $\displaystyle p$, for which one zero of the quadratic polynomial $\displaystyle \mathrm{px}^{2}-14 \mathrm{x}+8$ is $\displaystyle 6$ times the other.

Let the zeroes be \(\displaystyle \alpha\) and \(\displaystyle 6 \alpha\) Sum of zeroes \(\displaystyle =\alpha+6 \alpha=\frac{14}{\mathrm{p}}\) \(\displaystyle \Rightarrow \alpha=\frac{2}{\mathrm{p}}\) Product of zeroes \(\displaystyle =\alpha \times 6 \alpha=\frac{8}{p}\) \(\displaystyle \Rightarrow \quad 6 \times\left(\frac{2}{\mathrm{p}}\right)^{2}=\frac{8}{\mathrm{p}}\) \(\displaystyle \Rightarrow \quad \mathrm{p}=3\)