Clinton Montague

Developer, learner of things, functional programming enthusiast, hacker, and all round inquisitor.

Deriving the quadratic formula

November 1, 2014

This afternoon I decided to practice completing the square (I’m afraid that I’m rediscovering my inner nerd and am now studying for a math degree). After a few examples, I decided to go a bit more general and derive the quadratic formula. I guess that doing things like this is something which I’m going to have to get used to.

I figured that this is also a good excuse to practice my [latex size=”3″]\LaTeX[/latex] skills and write it out.

We want to find the general solution of the equation:

[latex size=4]ax^2 + bx + c = 0[/latex]

Start off by dividing through by [latex size=4]a[/latex]

[latex size=4]x^2 + \frac{b}{a}x + \frac{c}{a} = 0[/latex]

then complete the square

[latex size=4]\big(x + \frac{b}{2a}\big)^2 – \frac{b^2}{4a^2} + \frac{c}{a} = 0[/latex]

rearrange to leave the square on the left

[latex size=4]\big(x + \frac{b}{2a}\big)^2 = \frac{b^2}{4a^2} – \frac{c}{a}[/latex]

simplify the RHS

[latex size=4]\big(x + \frac{b}{2a}\big)^2 = \frac{b^2 – 4ac}{4a^2}[/latex]

take the square root of both sides

[latex size=4]\big(x + \frac{b}{2a}\big) = \pm \sqrt{\frac{b^2 – 4ac}{4a^2}}[/latex]

simplify the RHS

[latex size=4]\big(x + \frac{b}{2a}\big) = \pm \frac{\sqrt{b^2 – 4ac}}{2a}[/latex]

make [latex size=4]x[/latex] the subject

[latex size=4]x = – \frac{b}{2a} \pm \frac{\sqrt{b^2 – 4ac}}{2a}[/latex]

and, boom – the quadratic formula!

[latex size=4]\Rightarrow x = \frac{-b \pm \sqrt{b^2 – 4ac}}{2a}[/latex]

If I could go back in time and tell my younger self one thing, it would be to play with maths – don’t just do the homework. It’s all very well being able to follow a process (and essentially being a calculator), but knowing and understanding where something comes from unearths the true beauty of maths.