If you already know how to solve \(ax^{2} + bx + c = 0\) then this article is for you! Otherwise, it might be too confusing.

Ok, we know that the solution set to our quadratic can be found by the quadratic equation:

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

You can essentially derive this through a few ways, one of which is by completing the square.

There is another way that’s not really shown in Algebra classes and then as we progress through the math sequence, solving quadratics becomes too basic to bother with showing a proof or derivation on the solution method.

So, let’s explore. Let’s take our problem to solve

$$ax^{2} + bx + c = 0$$

Let’s guess that the solution is \(x = p + qi\) where \(p,q \in \mathbb{R}\) and \(i\) is the imaginary number. Our goal is to find what \(p\) and \(q\) are.

Substituting we have

$$a(p + qi)^{2} + b(p + qi) + c = 0$$ and now, a whole bunch of algebra. Let’s gooooooo!

$$a(p^{2}-q^{2}+2pqi) + bp + bqi + c = 0$$ collecting terms we have

$$(ap^{2}-aq^{2}+bp+c) + (q(2ap+ b))i = 0$$ now, notice that both the real and imaginary component must be zero. If we focus on the imaginary term, we have either \(q = 0\) in which case \(x = p\) and we have done nothing to advance the problem, or \(2ap + b = 0\) in which case \(p = -\frac{b}{2a}\). Does this look familiar??

Substituting this value of \(p\) into the real part we have

$$a\Big(-\frac{b}{2a}\Big)^{2}-aq^{2}+b\Big(-\frac{b}{2a}\Big)+c = 0$$ and solving for \(q\) we have

$$a\Big(-\frac{b}{2a}\Big)^{2}+b\Big(-\frac{b}{2a}\Big)+c = aq^{2}$$ dividing by \(a\)

$$\Big(-\frac{b}{2a}\Big)^{2}-\Big(\frac{b^{2}}{2a^{2}}\Big)+\frac{c}{a} = q^{2}$$ getting common denominators

$$\Big(\frac{b^{2}}{4a^{2}}\Big)-\Big(\frac{2b^{2}}{4a^{2}}\Big)+\frac{4ac}{4a^{2}} = q^{2}$$ condensing

$$\frac{-b^{2}+4ac}{4a^{2}} = q^{2}$$ and taking square roots

$$\frac{\pm\sqrt{-b^{2}+4ac}}{2a} = q$$ and putting \(p\) and \(q\) together we have

$$x = p + qi = -\frac{b}{2a} \pm \frac{\sqrt{-b^{2}+4ac}}{2a}i$$ and one final clean up

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

Do you see how this is the quadratic formula? The only difference is that the discriminant which is normally taught as \(b^{2} – 4ac\) is now just \(-b^{2} + 4ac\) because of the multiplication by \(i\).

I hope you enjoyed this!