For what value of '$p$' does the equation $x^2 - 2px + \left(7p - 12\right) = 0$ have equal roots?
Given quadratic equation: $\;$ $x^2 - 2px + \left(7p - 12\right) = 0$
Comparing with the standard equation $\;$ $ax^2 + bx + c = 0$ $\;$ gives
$a = 1$, $\;$ $b = -2 p$, $\;$ $c = 7p - 12$
A quadratic equation has equal roots when Discriminant $= \Delta = b^2 - 4ac = 0$
Here, discriminant $\;$ $\Delta = \left(-2p\right)^2 - 4 \times 1 \times \left(7p - 12\right)$
i.e. $\;$ $\Delta = 4 \left(p^2 - 7 p + 12\right)$
$\therefore \;$ For equal roots,
$4 \left(p^2 - 7p + 12\right) = 0$
i.e. $\;$ $p^2 - 7p + 12 = 0$
i.e. $\;$ $p^2 - 3p - 4p + 12 = 0$
i.e. $\;$ $p \left(p - 3\right) - 4 \left(p - 3\right) = 0$
i.e. $\;$ $\left(p - 3\right) \left(p - 4\right) = 0$
$\implies$ $p = 3$ $\;$ or $\;$ $p = 4$
$\therefore \;$ The given quadratic equation has equal roots when $p = 3$ or $p = 4$.