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Rational Roots Quadratic

Problem

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.

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

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