A polynomial intersecting the x-axis while not intersecting the x-axis?(Complex Numbers)

Question

I know three questions (that gained momentum) that have been posted asking a question which seems the same, but answers to none of them answer the following very well.

Please jump to point 2 & 3 for immediate addressing to the problem.

1. Knowledge that I currently have: I was introduced to imaginary numbers a while back where just the square root of -1 seemed to be called iota. This made a whole lot of 'things' definable/analyze-able. I have currently studied the two dimensional complex plane, the rotation, Euler form, defining locus/area which is indefinitely better than an algebraic equation/expression, etc.

2. My Question: However, I have failed to grasp that how imaginary numbers even make sense. Let's take an example, let there be a quadratic polynomial not intersecting the x-axis. However, it does with two imaginary numbers. How does it intersect the x-axis when the equation of the quadratic dictates that it doesn't intersect the x-axis (discriminant<0)?

3. My current understanding:Several sites led me to the conclusion that imaginary-axis acts as the 'z' axis and gets the quadratic to intersect with the x-axis somehow. This pushes the curve form a 2-D to a 3-D curve existing in the complex space when we start feeding complex values into the function with the x-axis as the intersection the the two, Cartesian and Complex plane. This 3-D structure now intersects somewhere on the complex plane with the x-axis giving us the complex roots of the quadratic.

I however fail to understand that how this curve transitions from the Cartesian plane to the complex space while being continuous. Does it transform to a surface instead of a curve? If it transforms to a surface then it would have infinite roots with the x-axis(as seen in the video mentioned in the footnote)

I understand that the complex numbers were 'invented' rather than discovered (I'm saying this because other numbers; irrational, rational, negatives actually mean something, hence discovered).

Thank you very much for your valuable time. I really appreciate it.

FOOTNOTE : I have been to the following questions/sites :

1

2

3

4

5 This video here is very good

6

EDIT: The question was marked as a duplicate though it did not in a way adress the same question as the question that was marked as 'original'. I have therefore, changed the title and content to make the question as precise and understandable as possible.

Answer 1

Your problem seems to be that you are trying to use the same intuition that you use for real numbers.

How does it intersect the x-axis when the equation of the quadratic dictates that it doesn't intersect the x-axis

It doesn't intersect the $x$-axis: the $x$-axis consists of real numbers. However, there are solutions that are complex numbers. These are not on the $x$-axis, because they are not real numbers.

Answer 2

A number system where we can do algebra needs to have addition, multiplication, subtraction and division and they need to behave well (for example $ab=0$ needs to imply $a=0$ or $b=0$, etc...). Such a system is called a "field". One such field is $\Bbb Q$ the rational numbers. Another is $\Bbb R$ the real numbers. Yet another is $\Bbb C$. And in fact $\Bbb Q\subseteq \Bbb R\subseteq \Bbb C$. It's just a set with four operations (two operations and their inverses really). Even though you cannot find $\Bbb C$ in your back yard, it's still a legitimate field that contains $\Bbb R$ and as such is a valid place to do algebra.

Now it turns out even if all you care about is $\Bbb Q$ it can be extremely useful to have the elbow room you get from thinking of $\Bbb Q$ as a subset of $\Bbb R$. And similarly in some cases (many cases) it's useful to have the additional elbow room you get from $\Bbb C$. So for example, Fermat's Last Theorem is a theorem about $\Bbb Q$ but if you didn't give yourself the elbow room of embedding $\Bbb Q$ into $\Bbb C$ then you would never be able to prove FLT.

Answer 3

Anyway did you ever see an irrational number?

This being said, when one asserts the parabola, say $y=x^2+x+1$ intersects the $x$-axis, one assumes $x$ and $y$ are complex numbers, i.e; the curve lives in $\mathbf C^2$, not in $\mathbf R^2$.

It is an algebraic variety of dimension $1$ over $\mathbf C$. In other words, from the real point of view, it is a surface embedded in a space of dimension $4$.