Explanation

(b)
Given, $\alpha^{2} - 5\alpha + 3 = 0$ and $\beta^{2} - 5\beta + 3 = 0$
$\Rightarrow \ \ \alpha = \frac{5 \pm \sqrt{13}}{2}$ and $\beta = \frac{5 \pm \sqrt{13}}{2}$
Since, $\alpha \neq \beta$
$\therefore\alpha = \frac{5 + \sqrt{13}}{2}$ and $\beta = \frac{5 - \sqrt{13}}{2}$
$\ \ \alpha = \frac{5 - \sqrt{13}}{2}$ and $\beta = \frac{5 + \sqrt{13}}{2}$
Now, $\alpha^{2} + \beta^{2} = \frac{50 + 26}{4} = 19$
And $\alpha\beta = \frac{1}{4}(25 - 13) = 3$
$\therefore$ Required equation is
$x^{2} - x\left( \frac{\alpha}{\beta} + \frac{\beta}{\alpha} \right) + \frac{\alpha\beta}{\alpha\beta} = 0$
$\Rightarrow \ \ x^{2} - x\left( \frac{\alpha^{2} + \beta^{2}}{\alpha\beta} \right) + 1 = 0$
$\Rightarrow \ \ 3x^{2} - 19x + 3 = 0$