Explanation

(c)
Since, $\alpha\ ,\ \beta$ are the roots of the equation $x^{2} + px + q = 0$, then
$\alpha + \beta = p,\ \ \alpha\beta = q$ ...(i)
and $\alpha^{4},\ \beta^{4}$ are the roots of $x^{2} - xr + s = 0$.
Then, $\alpha^{4} + \beta^{4} = r$ …(ii)
and $\alpha^{4}\beta^{4} = s$
If $D$ is discriminant of the equation $x^{2} - 4qx + 2q^{2} - r = 0$,
Then $D = 16q^{2} - 4\left( 2q^{2} - r \right) = 8q^{2} + 4r$
$= 8\alpha^{2}\beta^{2} + 4(\alpha^{4} + \beta^{4})$ [from Eqs. (i) and (ii)]
$= 4\left( \alpha^{2}\beta^{2} \right)^{2} \geq 0$
Hence, the equation $x^{2} - 4qx + 2q^{2} - r = 0$ has always two real roots.