The Sentimental Formula

I’m going to say something that may confuse quite a few people, Maths, at least for me, is sentimental and nostalgic. I can remember time back in my life that are important to me and I link them with Maths problems or topics. I even to some extent link people with topics and problems. One of those things is the proof of the Quadratic Formula.

The proof is one that involved minimal higher level Mathematics, but at the time I had just turned 16 and was working in the library. I was given a piece of paper from a friend, who I knew was also a Mathematician, well she was, as I was, teetering between a Physicist and a Mathematician. This proof also showed me and taught me more about how useful generality is in Maths.

We are going to need to do something called completing the square, this a way of writing: $ax^2 \,+\, bx \,+\, c = k(x \,+\, m)^2 \,+\, h$. This will be invaluable because from there we can subtract terms and square root. We can multiply out both sides of the above equation and then solve a system of equations in order to find our $k, m$ and $h$. Doing that will yield the following: $ax^2 + bx + c = kx^2 -2kmx + m^2k + h$. Now we can do something called equating coefficients, this will produce a system of equations to solve, which yields:

\begin{align*} a &= k \\ b &= \,-\,2km \\ c &= m^2k + h \end{align*}

Hence solving:

\begin{align*} k &= a \\ m &= \frac{b}{-2a} \\ h &= \frac{4ac \,-\, b^2}{4a} \end{align*}

We can now plug in and do the easy bit, the fiddling!

\begin{align*}a\left( x \, + \, \frac{b}{2a}\right)^2 + \frac{4ac \,-\, b^2}{4a} &= 0\\ \left( x \, + \, \frac{b}{2a}\right)^2 + \frac{4ac \,-\, b^2}{4a^2} &= 0 && \text{Dividing by $a$}\\ \left( x \, + \, \frac{b}{2a}\right)^2 &= \frac{b^2 \,-\, 4ac}{4a^2} \\ \end{align*}

I remember this bit ever so clearly, she asked me whether this looked familiar and I am to ask you the same, does it?

\begin{align*}\left( x \, + \, \frac{b}{2a}\right)^2 &= \frac{b^2 \,-\, 4ac}{4a^2} \\ \left( x \, + \, \frac{b}{2a}\right) &= \frac{\pm\sqrt{b^2 \,-\, 4ac}}{2a} \\ \therefore x &= \frac{-b\pm\sqrt{b^2 \,-\, 4ac}}{2a}\end{align*}

We have proved it, it was that simple. I look back extremely fondly of that time, as many people say a simpler time. Given the current climate, I feel that it is one of the many times where we will reflect and I know I have reflected a lot over the past weeks and feel that I will continue to.

I shall leave you with a proverb in her, and soon to be mine as well, favorite language.

‘It can be understood but not expressed in words’

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