Rational Roots Quadratic


In the quadratic equation $ax^2 + bx + c = 0$, the coefficients $a$, $b$, $c$ are non-zero integers.

Let $b = -5$. By making $a = 2$ and $c = 3$, the equation $2x^2 - 5x + 3 = 0$ has rational roots. But what is most remarkable is that it is possible to interchange these coefficients in any order and the quadratic will still have rational roots.

Suppose that $b$ is chosen at random. Prove that there always exist coefficients $a$ and $c$ that will produce rational roots. Moreover, once determined, no matter how these three coefficients are shuffled, the quadratic equation will still yield rational roots.


We shall prove this in two different ways.

Proof 1:
Although the first proof is elegant and provides a method for determining one set of values of $a$ and $c$, given $b$, it tells us nothing about the true nature of the problem, nor does it reveal that there are infinitely many different sets of values of $a$ and $c$ that can be determined for any given $b$.

It can be verified that the quadratic equation, $2x^2 + x - 3 = 0$ has rational roots and every arrangement of coefficients will yield rational roots. But the important observation is to note that any integral multiple of this "base" equation, with $b = 1$, will lead to another quadratic with rational roots for every arrangement. For example, if we multiply by 7, we get $14x^2 + 7x - 21 = 0$. This is equivalent to making $b = 7$ and determining that $a = 14$ and $c = -21$ produce a quadratic with rational roots for every arrrangement of the coefficients.

Proof 2:
This proof is perhaps the most revealling and makes use of the fact that the discriminant, $b^2 - 4ac$, must be the square of a rational if the roots of the equation are to be rational. In fact because all the coefficients are integer we can go further by saying that the discriminant must be square.

We also note that interchanging the positions of $a$ and $c$ has not effect on the rationality of the discriminant. Hence we only need consider the three cases of $a$, $b$, and $c$ being the coefficient of the $x$ term in the general quadratic. We shall initially consider the cases where $b$ and $c$ are the coefficient of $x$.

Let $b^2 - 4ac = r^2$ and $c^2 - 4ab = s^2$. We need to show that $r$ and $s$ are both integer.

$\begin{align}\therefore r^2 - s^2 &= b^2 - c^2 + 4ab - 4ac\\(r + s)(r - s) &= (b - c)(b + c + 4a)\end{align}$

Let $r + s = b + c + 4a$ and $r - s = b - c$.

Adding both equations we get $2r = 2b + 4a \implies r = b + 2a$, and subtracting gives, $2s = 2c + 4a \implies s = c + 2a$.

In other words, we have shown that $r$ and $s$ are both integer.

We must now show that the third discriminant, $a^2 - 4bc$, is also square.

Now $r^2 = (b + 2a)^2 = b^2 + 4ab + 4a^2 = b^2 - 4ac \implies 4a^2 + 4ab + 4ac = 0$

Similarly, $s^2 = (c + 2a)^2 = c^2 + 4ac + 4a^2 = c^2 - 4ab \implies 4a^2 + 4ab + 4ac = 0$

As $a \ne 0$, from $4a(a + b + c) = 0$ we deduce that $a + b + c = 0$.

As $a = -(b + c)$, squaring both sides gives, $a^2 = b^2 + c^2 + 2bc$. Subtracting $4bc$ from both sides:

$a^2 - 4bc = b^2 + c^2 -2bc = (b - c)^2$   QED

Is it possible that a quadratic equation exists for which any combination of the coefficients will yield rational roots and $a$ + $b$ + $c$ not equal 0?

Problem ID: 274 (21 Apr 2006)     Difficulty: 3 Star

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