A friend of mine gave me this problem
A quadratic of the form $ax^2+bx+c=0$ has its discriminant as one of its roots. What is the maximum possible value of $ab$?
I tried Vieta's, Quadratic Formula, even plugging it in, always ending up with complicated expressions that have, as far as I can tell, nothing to do with $ab$.
Is there another method of doing this? Am I missing something obvious? My friend said it shouldn't be too hard.
Thanks!
Best Answer
You have: $\Delta=\frac{-b-\sqrt {\Delta}}{2a}\,\vee \,\Delta =\frac{-b+\sqrt {\Delta}}{2a}$
Let's write this statement shorter as follows:
$$-b\pm\sqrt {\Delta}=2a\Delta$$
Let $\sqrt {\Delta}=u\ge 0$, then you get
$$ \begin{aligned}&2au^2\pm u+b=0\\ \implies &\Delta_u=1-8ab\ge 0\\ \implies &1\ge 8ab\\ \implies &ab\leq\frac 18.\end{aligned} $$
Explanatory note: Because $u$ is real (assuming $a,b,c$ are real), the discriminant of the quadratic $\Delta_u=1-8ab$ must be non-negative; therefore $\max\{ab\}=\dfrac 18.$