Introduction

In this very short blog post I want to go through the derivation of the quadratic equation as a simple teaching exercise. As most of us are aware from high school, when solving for the solutions to the following equation $$ ax^2 + bx + c = 0,\qquad a \neq 0 $$ one can use the quadratic equation, which has the general form. $$ x_{1, 2} = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} $$

Derivation

Initial Setup

To derive this equation one can use the following steps, $$ ax^2 + bx + c = 0 \implies ax^2 + bx =-c $$ As $a \neq 0$ we can divide through to get $$ x^2 + \frac{b}{a}x = -\frac{c}{a} $$

Completing the Square

Another fundamental piece of mathematics knowledge the most readers will be familier with is “completing the square”.

The general technique to complete the square is shown below $$ ax^2 + bx + c \implies \left( ax + \frac{b}{2}\right)^2 + c - \frac{b^2}{4} $$

Finishing Up

Using this knowledge we can now complete and solve the quadratic equation $$ x^2 + \frac{b}{a}x = -\frac{c}{a} \implies \left(x + \frac{b}{2a}\right)^2 - \frac{b^2}{4a^2} = -\frac{c}{a} $$

$$ \left(x + \frac{b}{2a}\right)^2 = \frac{b^2}{4a^2}-\frac{c}{a} $$

$$ x + \frac{b}{2a}= \pm\sqrt{\frac{b^2}{4a^2}-\frac{c}{a}} $$

$$ x = - \frac{b}{2a} \pm\sqrt{\frac{b^2}{4a^2}-\frac{c}{a}} $$ $$ x = - \frac{b}{2a} \pm\sqrt{\frac{b^2}{4a^2}-\frac{c \cdot 4a }{4a^2}} $$ $$ x = - \frac{b}{2a} \pm\frac{\sqrt{b^2-4ac}}{2a} $$

Which is the required result. $$\blacksquare$$