Deriving the Quadratic Formula

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.

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