Show that the equation $x^2 + px + 2p = 2x + 8$ has real roots for all real values of $p$.

1. Rearrange the equation into the standard quadratic form $ax^2 + bx + c = 0$.

$x^2 + px - 2x + 2p - 8 = 0$
$x^2 + (p - 2)x + (2p - 8) = 0$

2. Formulate and simplify the discriminant.

$\Delta = b^2 - 4ac$
$\Delta = (p - 2)^2 - 4(1)(2p - 8)$
$\Delta = p^2 - 4p + 4 - 8p + 32$
$\Delta = p^2 - 12p + 36$

3. Factorise the discriminant and conclude.

$\Delta = (p - 6)^2$
Since $(p - 6)^2 \ge 0$ for all real values of $p$, the discriminant is always non-negative ($\Delta \ge 0$).
Therefore, the equation has real roots for all real values of $p$.