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

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Step 1: $px(x - 2) + 6 = 0 \Rightarrow px^2 - 2px + 6 = 0$. Here $a = p$, $b = -2p$, $c = 6$.
Step 2: For equal roots: $D = 0 \Rightarrow b^2 - 4ac = 0 \Rightarrow 4p^2 - 24p = 0 \Rightarrow 4p(p - 6) = 0$
Step 3: $p = 0$ or $p = 6$. Since $p = 0$ is rejected (equation won't be quadratic), $	herefore p = 6$.

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