Tuesday, July 7, 2015

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

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

Some students would find that this problem very confusing, as they have been told countless time that $b^2-4ac$  is actually a discriminant for the quadratic equation $ax^2+bx+c=0$.

In this instance, $b^2-4ac$ is both the discriminant and the root, they would think instantly something that looks like the following:

$ax^2+bx+c=0$

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

Some students would go in hassle to expand the equation and they would get:

$a(b^4-8ab^2c+16a^2c^2)+b^3-4abc+c=0$

$ab^4-8a^2b^2c+16a^3c^2+b^3-4abc+c=0$

This transformation of the equation has turned into a giant mess, where $ab$ exists in $ab^4,\,8a^2b^2c,\,4abc$. Remember that our target is to find for the maximum of $ab$. It's unwise to stick and continue working with this equation to get $ab$ from it.

Now, what if we don't expand the equation after the replacement of $x$ by $b^2-4ac$ has been made?

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

Nope, after looking at the above equation without doing anything to it, it's not hard to see that there is practically nothing we do have done to gain something useful from there.

But some fact is always true, i.e. if $b^2-4ac$ is real, then the discriminant from the quadratic equation in $b^2-4ac$ must be greater than or equal to zero. And our target is to look for the maximum of $ab$.

The discriminant for $a(b^2-4ac)^2+b(b^2-4ac)+c=0$ is $b^2-4(a)(c)$. Aww, this is exactly the given root of $ax^2+bx+c=0$, it seems as though it has nothing to do with what we desire, to look for the maximum of $ab$.

In other words, we need to get rid of the constant $c$ in the quadratic equation and we don't want to include it in the

1. coefficient of $(b^2-4ac)^2$,

2. coefficient of $(b^2-4ac)$, and also

3. in the constant.

Wait a minute, we can use the quadratic formula to solve the equation for $x$ in the given quadratic equation!

$x=\dfrac{-b\pm \sqrt{b^2-4ac}}{2a}$

Replace $x$ by $b^2-4ac$ we see that:

$b^2-4ac=\dfrac{-b\pm \sqrt{b^2-4ac}}{2a}$

Rearrange the equation we obtain:

$2a(b^2-4ac)=-b\pm \sqrt{b^2-4ac}$

$2a(b^2-4ac)+b=\pm \sqrt{b^2-4ac}$

$2a(b^2-4ac)\mp \sqrt{b^2-4ac}+b=0$

Arrange it as another quadratic equation in $\sqrt{b^2-4ac}$, we get:

$2a\sqrt{b^2-4ac}^2\mp \sqrt{b^2-4ac}+b=0$

Ah! We're definitely heading in the right direction, as the coefficients of  $\sqrt{b^2-4ac}^2$ and $\sqrt{b^2-4ac}$ don't have the $c$ in them, also, the constant is a $b$!

Since we're told $b^2-4ac$ is real, so is $\sqrt{b^2-4ac}$, then the discriminant of the above quadratic equation in $\sqrt{b^2-4ac}$ must be greater than or equal to zero, i.e.

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

$1-8ab\ge 0$

$8ab\le 1$

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

and we're hence done.