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?
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