Without solving the equation, find the value of $p$ for which the equation $4x^2 + \left(p - 2\right)x + 1 = 0$ has equal roots.
Given quadratic equation: $\;$ $4x^2 + \left(p -2\right)x + 1 = 0$
Comparing with the standard quadratic equation $\;$ $ax^2 + bx + c = 0$ $\;$ gives
$a = 4, \; b = p - 2, \; c = 1$
The given quadratic equation has equal roots when discriminant $= \Delta = b^2 - 4ac = 0$
Here, $\;$ $\Delta = \left(p - 2\right)^2 - 4 \times 4 \times 1$
$\therefore \;$ We have for equal roots,
$\left(p - 2\right)^2 - 16 = 0$
i.e. $\;$ $p - 2 = \sqrt{16} = \pm 4$
i.e. $\;$ $p = 6$ $\;$ or $\;$ $p = -2$