Proving $b^2-4ac \leq \frac{1}{8}$

[Albert1]

$b^2-4ac$ is a real root of equation :$ax^2+bx+c=0,\,\, (a\neq 0)$

prove :$ab\leq \dfrac {1}{8}$

 

[anemone]

My solution:

Spoiler

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 simplify we get:

$-8ab+1\ge 0$

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

 

[Albert1]

very good you got it !
here is my solution :

Spoiler

$ax^2+bx+c=0---(1)$
solution of (1):$x=\dfrac {-b\pm\sqrt {b^2-4ac}}{2a}=b^2-4ac$
let :$\sqrt{b^2-4ac}=y----(*)$
from(*) we create two new equtions
$2ay^2\pm y+b=0----(**)$
since (**) have real solutions
$\therefore 1-8ab\geq 0$, or $ab\leq \dfrac {1}{8}$