Written by Colin+ in quadratics, there's more than one way to do it.

Factorising a quadratic? It's nice when it comes off, but there's a lot of guesswork, and no guarantee it even factorises.

Completing the square? Who has time for all that algebra?

And as for the quadratic formula, or your clever calculator methods: honestly, what are you, an engineer?

There is another way. A way passed from cowboy to cowboy1 for generations. A method that requires no guesswork and can be done without a calculator, and with (comparatively) little algebra.

Dear readers, I present to you, **Cowboy Completing The Square**.

I find it helps to have an example in mind, so suppose you have something like $x^2 - 4x - 8 = 0$.

We're going to rely on three facts:

- The properties of a quadratic graph means that the roots are equally spaced either side of the line of symmetry, and can be written in the form $(s-d)$ and $(s+d)$, where $x=s$ is the line of symmetry and $d$ is the distance away from it.
- The roots of a quadratic equation in the form $ax^2 + bx + c$ sum to $-\frac{b}{a}$.
- The product of the roots of a quadratic equation is $\frac{c}{a}$.

The sum of the roots is $(s-d)+(s+d) = 2s$, but we know that's also $-\frac{b}{a}$.

That tells us, for this example, that $2s = 4$ and $s=2$. In general, $s = -\frac{b}{2a}$, which is something I'm sure you've seen before.

The product of the roots is $(s-d)(s+d) = s^2 - d^2$, but we know that it's also $\frac{c}{a}$.

That gives us the equation $4 - d^2 = -8$, and we get $d^2 = 12$. We'd write that as $d=2\sqrt{3}$, of course.

So, knowing $s$ and $d$ means we can write down the solutions: $x = 2 + 2\sqrt{3}$ or $x = 2 - 2 \sqrt{3}$.

We can also use this method to factorise, as long as we take care about the $a$ when it's not zero. (For the example above, we can write the whole thing as $(x - (2+2\sqrt{3}))(x+(2-2\sqrt{3}))$ and be done with it - but when $a \ne 0$, we need to multiply everything by $a$ at the end.)

For example, given $6x^2 - 5x - 6$, we would immediately find $s= \frac{5}{12}$ and calculate $\br{\frac{5}{12}}^2 - d^2 = -1$. That gives $d^2 = \frac{169}{144}$ and $d=\frac{13}{12}$.

Our *zeros* are thus $\frac{3}{2}$ and $-\frac{2}{3}$, but the factorisation isn't $\br{x-\frac{3}{2}}\br{x+\frac{2}{3}}$, it's six times that. Multiplying the first bracket by two and the second by three gives $\br{2x-3}\br{3x+2}$, which does indeed expand to the original expression.

I think this is a lovely method, and one you can use mentally with only a little bit of practice. (I can't easily complete the square formally in my head, but cowboy-style? Yee-hah!)

- The historical accuracy of this statement is disputed. [↩]