Find the value of 'p' for which the quadratic equation $\displaystyle \mathrm{px}(\mathrm{x}-2)+6=0$ has two equal real roots.

\[\begin{gathered} p x(x-2)+6=0 \quad \Rightarrow p x^{2}-2 p x+6=0 \\ a=p, b=-2 p, c=6 \end{gathered} \] Quadratic equation has equal roots, \(\displaystyle \therefore \mathrm{D}=0\) \[\begin{aligned} & \quad \mathrm{b}^{2}-4 \mathrm{ac}=0 \quad \Rightarrow 4 \mathrm{p}^{2}-24 \mathrm{p}=0 \\ & \quad 4 \mathrm{p}(\mathrm{p}-6)=0 \\ & p=0, p=6 \\ & p=0 \text { rejected } \therefore p=6 \end{aligned} \] This section comprises of Long Answer (LA) type questions of $\displaystyle 5$ marks each.