Explanation

(b)
The roots of the equation $ax^{2} + \ bx + c = 0$ are given by
$x = \frac{- b \pm \sqrt{b^{2} - 4ac}}{2a}$
(i) Let $b^{2} - 4ac > 0,\ b > 0$
Now, if $a > 0,\ c > 0,\ b^{2} - 4ac < b^{2}$
$\Rightarrow$ The roots are negative.
(ii) Let $b^{2} - 4ac < 0,$ then the roots are given by
$x = \frac{- b \pm i\sqrt{(4ac - b^{2})}}{2a}$ ($i = \sqrt{- 1})$
Which are imaginary and have negative part. $(\because b > 0)$
$\therefore$ In each case the root have negative real part.