<s>[INST] Condition in terms of b and a if $ax^2+bx+c=0$ has two consecutive odd positive integers as roots 
The roots of the equation$$ax^2+bx+c=0$$, where $a \geq 0$, are two consecutive odd positive integers, then 

(A) $|b|\leq 4a$
(B) $|b|\geq 4a$
(C) $|b|=2a$
(D) None of these
My attempt
Let p and q be the roots then if they are consecutive positive integers (q>p) then $$ pq=\frac{c}{a} \geq 0$$
So, $$c \geq 0$$ and $$q-p=2$$
So, $$\frac{\sqrt{b^2-4ac}}{a}=2$$
So,$$|b|>2a$$
$(Since, a>0,c>0)$
But I know that 4ac should be taken in consideration since its not equal to zero. But I don't know how to use it. 
Any hints and suggestions are welcome!
 [/INST] Since we are given that there are two distinct roots and that $a \ge 0$, we must have
$a > 0, \tag 0$
since otherwise (i.e., with $a = 0$), the "quadratic" $bx + c$ has at most one root.
Let
$n \ge 0, \tag 1$
$r = 2n + 1, \tag 2$
$s = 2n + 3; \tag 3$
suppose for the moment 
$a = 1; \tag 4$
then
$(x - r)(x - s) = (x - (2n + 1))(x - (2n + 3)) = x^2 -(4n + 4)x + (2n + 1)(2n + 3) = x^2 -(4n + 4)x + (4n^2 + 8n + 3) = 0; \tag 5$
here we have
$b = -(4n + 4), \tag 6$
whence
$\vert b \vert = 4n + 4 \ge 4 = 4a; \tag 7$
thus (B) binds when $a = 1$; now if
$a \ne 1, \tag 8$
the quadratic of the form
$ax^2 + bx + c = a(x^2 + \dfrac{b}{a} x + \dfrac{c}{a}) \tag 9$
has zeroes $2n + 1$, $2n + 3$ provided
$(x - (2n + 1))(x - (2n + 3)) = x^2 -(4n + 4)x + (4n^2 + 8n + 3) =  x^2 + \dfrac{b}{a} x + \dfrac{c}{a}; \tag{10}$
thus,
$\dfrac{b}{a} = -(4n + 4), \tag{11}$
or
$b = -(4n + 4)a, \tag{12}$
whence, with $a > 0$,
$\vert b \vert = (4n + 4)a \ge 4a, \tag{13}$
and we see that the correct choice is (B) here as well.
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