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Q.
If $a>\,0, b>\,0, c>\,0$, then both the roots of the equation $a x^{2}+b x+c=0$
Complex Numbers and Quadratic Equations
Solution:
The roots of the equations are given by
$x=\frac{-b \pm \sqrt{b^{2}-4 a c}}{2 a}$
(i) Let $b^{2}-4 a c>\,0, b>\,0$
Now, if $a>\,0, c>\,0, b^{-}-4 a c<\,b^{-}$
$\Rightarrow $ the roots are negative.
(ii) Let $b ^{2}-4 ac <\,0$, then the roots are given by
$x =\frac{- b \pm i \sqrt{\left(4 ac - b ^{2}\right)}}{2 a },( i =\sqrt{-1})$
which are imaginary and have negative real part
$[\because b>\,0]$
$\therefore $ In each case, the roots have negative real part