Explanation

(c)
Since, $\alpha,\ \beta$ and $\alpha - k,\ \beta - k$ are the roots of the equations $ax^{2} + bx + c = 0$ and $Ax^{2} + Bx + C = 0$ respectively.
$\Rightarrow \ \alpha + \beta = - \frac{b}{a},\ \alpha\beta = \frac{c}{a}$
and $\alpha + \beta - 2k = - \frac{B}{A},\ \ (\alpha - k)(\beta - k) = \frac{C}{A}$
Now, $(\alpha - \beta)^{2} = (\alpha + \beta)^{2} - 4\alpha\beta = \frac{(b^{2} - 4ac)}{a^{2}}$ ...(i)
Also, $\left\{ (\alpha - k) - (\beta - k) \right\}^{2}$
$= \left\{ (\alpha - k) + (\beta - k) \right\}^{2} - 4(\alpha - k)(\beta - k)$
$= \left( \ - \frac{B}{A} \right)^{2} - 4\left( \frac{C}{A} \right)$
$= \frac{B^{2} - 4AC}{A^{2}}$ …(ii)
From Eqs. (i) and (ii)
$\frac{(b^{2} - 4ac)}{a^{2}} = \frac{B^{2} - 4AC}{A^{2}}$
$\therefore\ \frac{B^{2} - 4AC}{b^{2} - 4ac} = \left( \frac{A}{a} \right)^{2}$