So where does the quadratic formula come from, anyway?

The formula gives us the solutions of a quadratic equation of the form $ax^2 + bx + c = 0$ It tells us that this equation is true when $x = \frac{-b\pm\sqrt{b^2 – 4ac}}{2a}$

But where did such a strange formula come from?

It comes from solving the quadratic equation for $$x$$, but that requires some substitution. We can’t directly solve an equation that contains the same variable to different powers. Technically, $$x^2$$ is a different variable than $$x$$.

Instead, we need to rewrite the quadratic equation into a form that only has $$x$$ one time. Consider the expression $$(dx + e)^2$$: This satisfies that condition. We can solve $$(dx + e)^2 – f = 0$$ in terms of $$x$$: $(dx + e)^2 – f = 0 \Rightarrow \\ (dx + e)^2 = f \Rightarrow \\ dx + e = \pm \sqrt{f} \Rightarrow \\ dx = -e \pm \sqrt{f} \Rightarrow \\ x = \frac{-e \pm \sqrt{f}}{d} = \frac{-e}{d} \pm \frac{\sqrt{f}}{d}$

This looks similar to the quadratic formula, but simpler. If we could find a way to rewrite $$d$$, $$e$$, and $$f$$ in terms of $$a$$, $$b$$, and $$c$$, we’d be all set.

And we can do that. Let’s assume that the two forms of the quadratic equation represent an identical function, that is, $\color{red}{a}x^2 + \color{blue}{b}x + \color{green}{c} = (dx + e)^2 – f$ If we expand the right hand side, we get $(dx + e)^2 – f = \color{red}{d^2}x^2 + \color{blue}{2de}x + \color{green}{e^2 – f}$ This means: $\color{red}{a = d^2} \\ \color{blue}{b = 2de} \\ \color{green}{c = e^2 – f}$

The first line is straightforward: $$d = \sqrt{a}$$ (we’ll assume that d is positive; we could still derive the formula without this assumption, but it’s more confusing).

The second line then becomes $$b = 2e\sqrt{a}$$, so $$e = \frac{b}{2\sqrt{a}}$$.

The third line then becomes $$c = \frac{b^2}{4a} – f$$, so $$f = \frac{b^2}{4a} – c = \frac{b^2 – 4ac}{4a}$$.

At this point, we can substitute each variable. Start with $$\color{red}{f}$$: $\frac{-e}{d} \pm \frac{\sqrt{\color{red}{f}}}{d} = \frac{-e}{d} \pm \frac{\sqrt{\color{red}{\frac{b^2 – 4ac}{4a}}}}{d} \\ = \frac{-e}{d} \pm \frac{\sqrt{\color{red}{b^2 – 4ac}}}{\color{red}{2}d\color{red}{\sqrt{a}}}$

Next, replace $$\color{red}{e}$$: $\frac{-\color{red}{e}}{d} \pm \frac{\sqrt{b^2 – 4ac}}{2d\sqrt{a}} = \frac{-\color{red}{\frac{b}{2\sqrt{a}}}}{d} \pm \frac{\sqrt{b^2 – 4ac}}{2d\sqrt{a}} \\ = \frac{-\color{red}{b}}{\color{red}{2}d\color{red}{\sqrt{a}}} \pm \frac{\sqrt{b^2 – 4ac}}{2d\sqrt{a}} = \frac{-b \pm \sqrt{b^2 – 4ac}}{2d\sqrt{a}}$

Finally, replace $$\color{red}{d}$$: $\frac{-b \pm \sqrt{b^2 – 4ac}}{2\color{red}{d}\sqrt{a}} = \frac{-b \pm \sqrt{b^2 – 4ac}}{2\color{red}{\sqrt{a}}\sqrt{a}} = \frac{-b \pm \sqrt{b^2 – 4ac}}{2a}$

Recall that we started with an equation involving $$d$$, $$e$$, and $$f$$ which represented the values of $$x$$ that made the equation true. That is, $x = \frac{-e \pm \sqrt{-f}}{d} = \frac{-b \pm \sqrt{b^2 – 4ac}}{2a}$ which is the quadratic formula in its typical form.