<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] Using the quadratic formula, we have that (using $p<q$ as the roots):
$$p+2=q$$
$$\frac{-b-\sqrt{b^2-4ac}}{2a}+2=\frac{-b+\sqrt{b^2-4ac}}{2a}$$
$$\frac{-b}{2a}-\frac{\sqrt{b^2-4ac}}{2a}+2=\frac{-b}{2a}+\frac{\sqrt{b^2-4ac}}{2a}$$
$$2=2\frac{\sqrt{b^2-4ac}}{2a}$$
$$2a=\sqrt{b^2-4ac}$$
$$4a^2=b^2-4ac$$
$$b^2=4a^2+4ac$$
$$\frac{b^2}{a^2}=4(1+\frac{c}{a})=4(1+pq)$$
$$\frac{|b|}{a}=2\sqrt{1+pq}$$
Then, given that $p,q$ are different odd integers, you can show that $pq\geq3$, so that $2\sqrt{1+pq}\geq2\sqrt{1+3}=4$
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