Ytukyg

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!

Simply, If $2n-1$ and $2n+1$ are the roots (with $nge 1$) 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 $|b|=4nage 4a$.

Using the quadratic formula, we have that (using $p<q$ as the roots):
$$p+2=q$$
$$frac{-b-sqrt{b^2-4ac}}{2a}+2=frac{-b+sqrt{b^2-4ac}}{2a}$$
$$frac{-b}{2a}-frac{sqrt{b^2-4ac}}{2a}+2=frac{-b}{2a}+frac{sqrt{b^2-4ac}}{2a}$$
$$2=2frac{sqrt{b^2-4ac}}{2a}$$
$$2a=sqrt{b^2-4ac}$$
$$4a^2=b^2-4ac$$
$$b^2=4a^2+4ac$$
$$frac{b^2}{a^2}=4(1+frac{c}{a})=4(1+pq)$$
$$frac{|b|}{a}=2sqrt{1+pq}$$
Then, given that $p,q$ are different odd integers, you can show that $pqgeq3$, so that $2sqrt{1+pq}geq2sqrt{1+3}=4$

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.

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.)