Theorem – Quadratic formula – Math-reference.net

Derivation of the quadratic equation.

Theorem

Given any equation of the form $f(x) = ax^2 + bx + c$ , the roots, or zeroes of $f$ are given by the formula,

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

Proof

Given an equation $f(x) = ax^2 +bx +c$ , to find the roots of $f$ is equivalent to finding all $x$ such that $ax^2 + bx + c = 0$ . We will also assume that $a \ne{0}$ . It follows that,

$\begin{aligned} ax^2 + bx + c &= 0 \\\Leftrightarrow x^2 + \frac{b}{a}x + \frac{c}{a} &=0 \\\Leftrightarrow x + (\frac{b}{2a})^2 +\frac{c}{a} - \frac{b^2}{(2a)^2} &= 0 \end{aligned}$

Now, rearranging this equation we obtain,

$\begin{aligned} \\\Leftrightarrow (x + \frac{b}{2a})^2 +\frac{c}{a} & = \frac{b^2}{4a^2} \\\Leftrightarrow (x + \frac{b}{2a})^2 & = \frac{b^2}{4a^2}-\frac{c}{a} \\\Leftrightarrow (x + \frac{b}{2a})^2 & = \frac{b^2}{4a^2}-\frac{4ac}{4a^2} \\\Leftrightarrow (x + \frac{b}{2a})^2 &= \frac{b^2 - 4ac}{4a^2} \\\Leftrightarrow x + \frac{b}{2a} & = \pm\sqrt{\frac{b^2 - 4ac}{4a^2}} \\\Leftrightarrow x + \frac{b}{2a} & = \pm\frac{\sqrt{b^2 - 4ac}}{2a} \\\Leftrightarrow x & = - \frac{b}{2a} \pm\frac{\sqrt{b^2 - 4ac}}{2a} \\\Leftrightarrow x & = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} ~\square \end{aligned}$

Thus, we have shown that, as long as $a \ne{0}$ , $a, b, c \in \mathbb{R}$ , and $\sqrt{b^2 - 4ac} \geq 0$ ,

$\displaystyle{ax^2 + bx + c = 0 \Leftrightarrow x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}}$

Notes

We can still find values of $x$ if these conditions are not met, for example if $a = 0$ , then $x = -c$ , since $ax^2 +bx +c=0$ reduces to $bx + c=0$ .

Furthermore, when $b^2 - 4ac < 0$ , we can still find solutions to these equations using complex numbers. The development of which can be found here.