Condition in terms of b and a if $ax^2+bx+c=0$ has two consecutive odd positive integers as roots

Question

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!

Answer 1

Simply, If $2n-1$ and $2n+1$ are the roots (with $n\ge 1$) then $$ax^2+bx+c=a(x-2n+1)(x-2n-1)=a((x-2n)^2-1)=a(x^2-4nx+4n^2-1) $$ so $b=-4na$ and hence $|b|=4na\ge 4a$.

Answer 2

Easy way to check:

Take a quadratic equation which has 1 and 3 as roots. $(x-1)(x-3)=x^2-4x+4x$, so you see that it is possible for $|b|=4|a|$, ruling out choice (C).

Do the problem again choosing the roots 3 and 5. $(x-3)(x-5)=x^2-8x+15$, so you can conclude that $|b|>=4a$, and (B) is correct. (If you continue choosing larger pairs of numbers as roots, $\frac{|b|}{a}$ will only increase.)

Answer 3

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 $2\sqrt{1+pq}\geq2\sqrt{1+3}=4$

Answer 4

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.