Bascially we are trying to find the roots of a quadratic equation, and 'apparently' there is a theorem for this, but every one that I have found so far mentions that the degree of the polynomial is directly related to the number of zeros (where X is an integer and Y is 0).. However, most of the theorems also mention complex numbers, which our math teacher does not want us to include in this theorem. Does anyone know of a way to find the number of real zereos using a theorem that does not involve complex numbers?
Note: The theorem also has to account for vertical shifts, which I don't see how that can be possible.
Best Answer
The quadratic functions are generally in the form $f(x)=Ax^2+Bx+C$. The zeroes of this function is then the solution to the equation $Ax^2+Bx+C=0$.$($where $A\ne 0)$
Hence we try to solve this equation.
If you try to solve the equation (just like you'd solve a linear equation), you'd run into trouble when you realize that the $x$s can't be as easily isolated like in the case with linear equations. So we use a different approach, what is called the "completion of squares".
If you are familiar with the method of "completing the squares" (and it isn't very difficult,) you'd know that our goal is to somehow "arrange" the equation in the form $(\text{linear binomial})^2+\text{constant term}$. When you arrange the equation in this way, isolating the $x$s is very easy. I'll not show you the entire process of completing the square for our given function but I presume that you are somewhat familiar with it (if not, I suggest you look it up on the internet).
The end result of our method will yield a very familiar expression, what we are used to calling as the "quadratic formula". You should end up with something like this: $$x=\frac{-B}{2\cdot A} \pm \frac{\sqrt{B^2-4\cdot A\cdot C}}{2\cdot A}$$ where $A$, $B$ and $C$ are the coeffients of our function $f(x)$.
The relation above should give you the zeroes of any quadratic function with coefficients $A$, $B$ and $C$ ( Also note that $A$, $B$, $C \in \mathbb R$). Note that this particular relation has many take-aways. First it hints you of the existence of $2$ zeroes (notice the $\pm$ sign in our relation). Second it kind of explains why $ A \ne 0$ (if $A=0$ then the whole relation goes meaningless, for when you divide any thing by $0 $; in our case $2 \times 0 $, the result is undefined.)
The third, and probably the most important one, is that it tells you when $x$ is a Real number or when it is not (i.e when it is a Complex number). Remember that while $A$, $B$ and $C$ must always be Real numbers, it isn't necessary for $x$ to be Real. It is a well known fact square roots of negative numbers are not Real number, instead, we call them the imaginary numbers $($ and the complex of Real and Imaginary numbers are called the Complex Numbers, usually in the form $a+b\mathbb i $ and $\mathbb i =\sqrt{-1})$.
So,
The geometric interpretation of these three cases is also pretty interesting and gives a deeper insight at what is going on (Although your question has already been answered, i suggest you keep reading, I'm almost done :)).
You should know that the zeroes of a function are those values of $x$ which results in $f(x)$ being equal to zero. Or, if you graph the function, It is that point on the graph(or that value for $x$ ), where the function crosses the x-axis. So, what do these three cases imply on the graph then?
And with that I have come to the end of my uber-long answer. (Phew!)
As far as your question is concerned, If you'd like to know if the zeroes are Real or not, check if $B^2 -4AC$ is positive, negative or zero, the nature of the roots (zeroes) should be apparent.
I bothered to write down such long an answer because I really wanted you to know what's going on behind the scenes. You might already know this stuff but still, I Hope it helps