Thursday, July 9, 2015

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:


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


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:







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:


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!

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