Explanation

(a)
Since, $\alpha\ and\ \beta$ are the roots of $ax^{2} + bx + c = 0$.
$\Rightarrow \ \alpha + \beta = - \frac{b}{a}\ and\ \alpha\beta = \frac{c}{a}$ ...(i)
If roots are $\ \alpha + \frac{1}{\beta},\beta + \frac{1}{\alpha}$, then
Sum of roots $= \left( \ \alpha + \frac{1}{\beta} \right) + \left( \ \beta + \frac{1}{\alpha} \right) = (\alpha + \beta) + \frac{\alpha + \beta}{\alpha\beta}$
$= \frac{- b}{ac}(a + c)$ [from Eq. (i)]
$= \alpha\beta + 1 + 1 + \frac{1}{\alpha\beta} = 2 + \frac{c}{a} + \frac{a}{c}$ [from Eq. (i)]
and product of roots $= \left( \ \alpha + \frac{1}{\beta} \right) + \left( \ \beta + \frac{1}{\alpha} \right)$
$= \frac{2ac + c^{2} + a^{2}}{ac} = \frac{(a + c)^{2}}{ac}$
Hence, required equation is given by
$x^{2} - (sum\ of\ roots)x +$(prouduct of roots)=0
$\Rightarrow x^{2} + \frac{b}{ac}(a + c)x + \frac{(a + c)^{2}}{ac} = 0$
$\Rightarrow acx^{2} + (a + c)bx + (a + c)^{2} = 0$