Explanation

(d)
Given, $\alpha + \beta = 4$ and $\alpha^{3} + \beta^{3} = 44$
$\Rightarrow \ \ \ (\alpha + \beta)^{2} - 3\alpha\beta(\alpha + \beta) = 44$
$\Rightarrow 64 - 44 = 12\alpha\beta\ \ \Rightarrow \ \ \ \alpha\beta = \frac{5}{3}$
$\therefore$ Required equation is
$x^{2} - (\alpha + \beta)x + \alpha\beta = 0\ \Rightarrow x^{2} - 4x + \frac{5}{3} = 0$
$\Rightarrow \ \ \ 3x^{2} - 12x + 5 = 0$