Connection between quadratic equations and complex numbers where $\Delta > 0$ (of the quadratic formula)

Question

I recently studied how the complex quadratic equation of $x^2-2x+5$ was solved:

$x^2-2x+5$">

Then I typed it in for the star of this question:

$$x^2-2x-5$$

And it used the quadratic formula, which I found boring. So I wondered: what will happen if I use the same method as the image above and just pretend it will have a complex solution. Will the imaginary numbers cancel out? What will happen?

So I did that, and my question is: how did this work so well? What is actually happening here? Is this a complex number thing, or just the fact that I'm splitting $x$ into two components?

Or did I just luck into the solution and actually have a subtle mistake?

Here are the steps I took.

$$x^2-2x-5=0$$

$$(a+bi)^2 - 2(a+bi) - 5 = 0$$

$$(a^2 + 2abi +b^2i^2) - 2a - 2bi - = 0$$

Reordering it to the 'template' of the image. (note: $b^2i^2 = -b^2$)

$$(a^2 - b^2 - 2a - 5) + i(-2b + 2ab) = 0$$

Alright, I'm on the right track as it is intuitive to me that only -5 should be the only changed value.

\begin{bmatrix} a^2 - b^2 - 2a - 5 = 0 \\ -2b + 2ab = 0 \end{bmatrix}

Here is where I venture of a little bit as my linear algebra is almost non-existent.

$$2ab - 2b = 0$$

$$2ab = 2b$$

$$2a = 2$$

$$a = 1$$

Since it's immediately visible what $b$ is in the same equation, I went to the other equation as I figured, it might have more information about $b$. We already have a believable value for $a$, so I subtitute it in there.

$$1^2 - 2*1 - b^2 - 5 = 0$$

$$-b^2 - 6 = 0$$

$$-6 = b^2$$

$$b = \sqrt{-6}$$

$$b = \sqrt{6}i \text{ or } b = -\sqrt{6}i$$

And substitute back for $x = a + bi$ we get:

$$x = 1 + (\pm\sqrt{6}i)i$$

$$x = 1 + (\pm\sqrt{6}i^2)$$

$$x = 1 + (\pm\sqrt{6}*-1)$$

$$x = 1 + (\pm\sqrt{6})$$

$$x = 1 \pm\sqrt{6}$$

Wait, that is correct (use the quadratic formula or something like SymPy to check, I use the website above called https://www.symbolab.com/solver/).

How is this possible? How is it possible that if I model this equation with complex numbers, I also get a valid solution? When isn't this possible?

Answer 1

You supposed that the solution had the form $a+bi$ but you permitted $a,b\in\mathbb{C}$. While this will yield a complex number (since they are closed under addition and multiplication), it is not quite the general form of a complex number, where we must have $a,b\in\mathbb{R}$.

The reason why this works is because of the fundamental theorem of algebra. $a+bi$ is still some complex number, even if we permit $a,b\in\mathbb{C}$, and by the fundamental theorem of algebra, every non-constant single variable polynomial has at least one complex root. And the real numbers are a subset of the complex numbers, that is $\mathbb{R}\subset\mathbb{C}$. So even by presuming that your solutions are complex, it is perfectly possible to generate the real solutions, since all real numbers are complex.

Notice if you had used the standard form of a complex number, with $a,b\in\mathbb{R}$, you would have noted that $2ab-2b=0$ factors as $2b(a-1)=0$. So either $b=0$ or $a=1$. In the latter case, you would have found that $b$ would have to be nonreal, which contradicts your presumption that $b\in\mathbb{R}$. So then, you would be forced to have $b=0$ and $a^2-2a-5=0$, which gets us right back to where we started.

Answer 2

It's hard to figure out what kind of answer you are looking for because there are many possible ways to provide meaningful explanations. The whole reason the complex numbers got defined is to add more structure to the already existing real numbers. In more modern mathematical terms this is formally known as a field extension.

The way you already know the real numbers $\mathbb{R}$ to work, together with the $+$ and $\times$ operations, is formally called a "field" symbolically written as $(\mathbb{R}, +, \times)$. These operations still work the same way in the complex numbers, making it a field as well, symbolically written as $(\mathbb{C}, +, \times)$. Furthermore the real numbers are contained inside the complex numbers, by virtue of $a +bi \in \mathbb{R} \iff b = 0$. So naturally, since the $+$ and $\times$ operations work the same way in both fields, it is possible to solve real problems in the complex space.