# An unorthodox way of solving quadratic equations

This post concerns an unorthodox but logically correct technique for solving a quadratic equation via factoring. I showed this to some senior math majors as well as graduate students in mathematics; none of them had ever seen this before. Suppose that we want to solve $6x^2 - 13x - 5 = 0$

without using the quadratic formula. Trying to solve this by factoring looks like a pain in the neck, as there are several possibilities: $(x + \underline{\quad})(6x - \underline{\quad}) = 0$, $(x - \underline{\quad})(6x + \underline{\quad}) = 0$, $(2x + \underline{\quad})(3x - \underline{\quad}) = 0$,

or $(2x - \underline{\quad})(3x + \underline{\quad}) = 0$.

So instead, let’s replace the original equation with a new equation. I’ll get rid of the leading coefficient and multiply the constant term by the leading coefficient: $t^2 - 13t - (5)(6) = 0$, or $t^2 - 13t - 30 =0$.

This is a lot easier to factor: $(t - 15)(t+ 2) = 0$ $t = 15 \quad \hbox{or} \quad t = -2$

So, to solve for $x$, divide by the original leading coefficient, which was $6$: $x = 15/6 = 5/2 \quad \hbox{or} \quad x = -2/6 = -1/3$.

As you can check, those are indeed the roots of the original equation.

This technique always works if the quadratic polynomial has rational roots. But why does it work? I’ll give the answer after the thought bubble. The original quadratic equation was $6x^2 - 13x - 5 = 0$

Let’s make the substitution $x = t/6$: $6 \displaystyle \left( \frac{t}{6} \right)^2 - 13 \left( \frac{t}{6} \right) - 5 = 0$ $\displaystyle \frac{t^2}{6} - \frac{13t}{6} - 5 = 0$

Multiply both sides by $6$, and we get the transformed equation: $t^2 - 13t - 30 = 0$

Although I personally love this technique, I have mixed feelings about the pedagogical usefulness of this trick… mostly because, to students, it probably feels like exactly that: a trick to follow without any conceptual understanding. Perhaps this trick is best reserved for talented students who could use an enrichment activity in Algebra II.

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