## Friday, August 14, 2015

### Find the sum of all possible $a^3$, where $a$ is a rational figure.

Given that $a$ is rational and the equation $ax^2+(a+2)x+a-1=0$ has integer roots.

Find the sum of all possible $a^3$.

The solution will require deep and strategic thought but that doesn't mean this problem is impossible to solve.

If you know and are familiar with the manipulation trick that we could play on the rational number, we can see the solution pretty clearly.

But, let's say if you begin to attempt at the solution WITHOUT a plan, i.e. you just put the pen to paper and begin to write things, then you would head to nowhere nearer to the solution.

That is what the vast majority of the students would do:

Since we're told that the given quadratic equation $ax^2+(a+2)x+a-1=0$ has integer roots, we know we could "play" on its discriminant that the discriminant must be greater than zero, and is a perfect square.

\begin{align*}\text{Discriminant}&=(a+2)^2-4(a)(a-1)\\&=4+8a-3a^2\\&=\left(\left(\dfrac{4-2\sqrt{7}}{3}\right)-a\right)\left(\left(\dfrac{4+2\sqrt{7}}{3}\right)-a\right)\gt 0\end{align*}

This results in $\dfrac{4-2\sqrt{7}}{3} \lt a \lt \dfrac{4+2\sqrt{7}}{3}$.

We could not do anything to make use of the other info that the discriminant must be a perfect square, since $a$ is a rational number.

But, from the given quadratic equation, we could tell that the product of the integers root must be another integer, i.e. $\dfrac{a-1}{a}=1-\dfrac{1}{a}$ must be an integer. This literally means $-1\le a\le 1$.

Combining the two inequalities gives us:

$\dfrac{4-2\sqrt{7}}{3} \lt a \le 1$

But all that we know is that $a$ is a rational number, and that could yield many cases we have to consider to see if each case meet two conditions where $x$ is an integer and also $a$ is a rational number.

It couldn't be feasible to fully list out the seemingly endless possibilities, as $a$ could be $-\dfrac{1}{2},\,\dfrac{1}{4},\,\dfrac{3}{5}$ just to name a few.

Therefore, we need to think of other approach, and we must "get rid of" $a$, the rational figure so our work is more manageable and we won't have to check with endless cases to solve for this problem.

I hope you would begin to think deep and hard now on a cold and rational level, and I will lead you to the solution in the next blog post.