Question

If the roots of the equation $a{x}^2+bx+c=0,a\in R^{+}$ are two consecutive odd positive integers, then

• A. $|b|\le 4a$
• B. $|b|\ge 4a$
• C. $|b|\ge 2a$
• D. $|b|\le a$

Answer 1

Answer: (C)

$|b|\ge 2a$

Let $\alpha ,\beta$ be the roots$\Rightarrow \alpha +\beta =\frac{-b}{c};\alpha \beta =\frac{c}{a}$
$\alpha -\beta =2\Rightarrow \sqrt{{\left(\frac{b}{a}\right)}^{{}^2}-4\left(\frac{c}{a}\right)}=|\alpha -\beta |$
$\sqrt{\frac{b^2-4ac}{a^2}}=2.$
$b^2-4ac=4a^2$
$b^2=4a^2+4ac$
As both $\alpha$ and $\beta$ are positive, $ac$ is positive
$\Rightarrow b^2\ge 4a^2$
$\Rightarrow |b|\ge 2a$