Question

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

• A. $|b|\le 4a$
• B. $|b|\ge 4a$
• C. $|b|=2a$
• D. None of these

Answer 1

Answer: (B)

Let the roots be $\alpha$ and $\alpha +2$,
where $\alpha$ is an odd positive integer.
Then, $a{a}^2+b\alpha +c=0$
and $a(\alpha +2{)}^2+b(\alpha +2)+c=0$
$\Rightarrow a{a}^2+b\alpha +c+(4a\alpha$ $+4a+2b)=0$
$\Rightarrow 2a(1+\alpha )+b=0$[using (1)]
$\Rightarrow b=-2a(1+\alpha )$
$\Rightarrow b^2=4a^2(1+\alpha {)}^2$
$\Rightarrow b^2\ge 4a^2(1+1{)}^2$
$[\because \alpha \ge 1$ as $\alpha$ is odd positive integer $]$
$\Rightarrow b^2\ge 16a^2$
or $|b|\ge 4a$