Prove that $ab\leq \dfrac {1}{8}$ (Second Solution)

Given $b^2-4ac$ is a real root of equation :$ax^2+bx+c=0,\,\, (a\neq 0)$. Prove that $ab\leq \dfrac {1}{8}$.

Second solution is my solution: (I want to mention it here that the first solution is provided by a Taiwan friend of mine)

Multiply the equation of $ax^2+bx+c=0$ by $4a$, we get:

$4a^2x^2+4abx+4ac=0$(*)

We're told that $b^2-4ac$ is a real root of the equation $ax^2+bx+c=0$, this tells us:

1. $4ac$ is a real number.

2. And we can, in this case, substitute $x=b^2-4ac$ into the equation (*), to get:

$4a^2(b^2-4ac)^2+4ab(b^2-4ac)+4ac=0$

Rewrite the above equation as another quadratic equation in terms of $4ac$, we see that we have:

$(4a^2)(4ac)^2-(8a^2b^2+4ab-1)(4ac)+(4a^2b^4+4ab^3)=0$

Since $4ac$ must be a real number, so the discriminant of the above equation must be greater than or equal to zero, thus this yields:

$(-(8a^2b^2+4ab-1))^2-4(4a^2)(4a^2b^4+4ab^3)\ge 0$

Expanding and simplifying we get:

$-8ab+1\ge 0$

$\therefore ab\le \dfrac{1}{8}$

And admittedly, this solution is less elegant than the first solution, but it does provide us with another way of attacking the problem and digested the given info entirely differently.

I believe those who have been followed my posts know that I am a strong advocate in trying my very best to give people as many different ways as possible of looking at the mathematical problems.