Theory of Equations 1 Question 36
37. Let $a>0, b>0$ and $c>0$. Then, both the roots of the equation $a x^{2}+b x+c=0$
(1979, 1M)
(a) are real and negative
(b) have negative real parts
(c) have positive real parts
(d) None of the above
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Answer:
Correct Answer: 37. (b)
Solution:
- Since, $a, b, c>0$ and $a x^{2}+b x+c=0$
$\Rightarrow \quad x=\frac{-b}{2 a} \pm \frac{\sqrt{b^{2}-4 a c}}{2 a}$
Case I When $b^{2}-4 a c>0$
$\Rightarrow \quad x=\frac{-b}{2 a}-\frac{\sqrt{b^{2}-4 a c}}{2 a}$
and $\frac{-b}{2 a}+\frac{\sqrt{b^{2}-4 a c}}{2 a}$ both roots, are negative.
Case II When $b^{2}-4 a c=0$
$\Rightarrow x=\frac{-b}{2 a}$, i.e. both roots are equal and negative
Case III When $b^{2}-4 a c<0$
$ \Rightarrow \quad x=\frac{-b}{2 a} \pm i \frac{\sqrt{4 a c-b^{2}}}{2 a} $
have negative real part.
$\therefore$ From above discussion, both roots have negative real parts.
