Explanation

(a)
Since, $\sin{\alpha\ and\cos\alpha}$ are the roots of the equation $ax^{2} + bx + c = 0,$ then
$\sin{\alpha + \cos{\alpha = - \frac{b}{a}}\ }$ and $\sin{\alpha\ \cos{\alpha = \frac{c}{a}}}$
To eliminate $\alpha$, we get
$1 = \sin^{2}{\alpha + \cos^{2}\alpha}$
$\Rightarrow 1 = {(\sin{\alpha + \cos{\alpha)}\ }}^{2} - 2\sin{\alpha\ \cos\alpha}$
$\Rightarrow 1 = \frac{b^{2}}{a^{2}} = \frac{2c}{a}$
$\Rightarrow a^{2} - b^{2} + 2ac = 0$