Explanation

(c)
Here, $\alpha + \beta = - 5$ and $\alpha\beta = 4$
Now, $\frac{\alpha + 2}{3} + \frac{\beta + 2}{3} = \frac{\alpha + \beta + 4}{3} = \frac{- 5 + 4}{3} = \frac{- 1}{3}$
And $\left( \frac{\alpha + 2}{3} \right)\left( \frac{\beta + 2}{3} \right) = \frac{\alpha\beta + 2(\alpha + \beta) + 4}{9}$
$= \frac{4 + 2( - 5) + 4}{9} = \frac{- 2}{9}$
Required equation is
$x^{2}$-(sum of roots)$x +$products of roots=0
$\therefore x^{2} + \frac{1}{3}x - \frac{2}{9} = 0\ \Rightarrow 9x^{2} + 3x - 2 = 0$