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
Assertion and Reason
For the following question, choose the correct answer from the codes (a), (b), (c) and (d) defined as follows :
(a) Statement I is true, Statement II is also true; Statement II is the correct explanation of Statement I
(b) Statement I is true, Statement II is also true; Statement II is not the correct explanation of Statement I
(c) Statement I is true; Statement II is false
(d) Statement I is false; Statement II is true
Show Answer
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.
