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

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

Third solution:

Since our aim is targeting at reaching a quadratic equation in $b^2-4ac$ or even $\sqrt{b^2-4ac}$, where $c$ is absent from the coefficients of the square term, the linear term and also the constant, there is one more way to do that without solving the given quadratic equation in $x$ using the quadratic formula:

We're given:

$ax^2+bx+c=0$

We're told $b^2-4ac$ is a real root of it so we have:

$\displaystyle \color{yellow}\bbox[5px,purple]{a(b^2-4ac)^2+b(b^2-4ac)+c=0}$(*)

We need to kick $c$ off, that is feasible if we transform $c$ into either $b^2-4ac$ or $-(b^2-4ac)$ so we could group it to the second linear term of $b(b^2-4ac)$:

That means, we have to algebraically manipulate the equation (*) so we have:

$\displaystyle \color{yellow}\bbox[5px,purple]{a(b^2-4ac)^2+b(b^2-4ac)+c=0}$(*)

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

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

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

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

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

As you can see it now, the coefficients of $(b^2-4ac)^2$ and $b^2-4ac$ are free from $c$, so is with the constant:

$\displaystyle \color{yellow}\bbox[5px,purple]{4a^2}\color{black}(b^2-4ac)^2+\color{yellow}\bbox[5px,green]{(4ab-1)}\color{black}(b^2-4ac)+\color{black}\bbox[5px,orange]{b^2}\color{black}=0$

We can safely say it our loud now that since $b^2-4ac$, the discriminant to the above quadratic in $b^2-4ac$  must be greater than or equal to zero, i.e.

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

$16a^2b^2-8ab+1-16a^2b^2\ge 0$

$\cancel{16a^2b^2}-8ab+1-\cancel{16a^2b^2}\ge 0$

$1\ge 8ab$

Therefore we get:

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

and we're done!