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.

  • 2
    $\begingroup$ Use the discriminant of the quadratic polynomial to find the number of (real) roots. $\endgroup$
    – JavaMan
    Feb 24, 2012 at 17:13
  • $\begingroup$ I don't understand what you mean by "account for vertical shifts". The discriminant of the quadratic polynomial $f(x) = ax^2 + bx + c$ is $b^2 - 4ac$. If it is positive, the quadratic polynomial has two real roots. If it is zero, the polynomial has one real root (with multiplicity two), and if it is negative, the polynomial has two complex (and hence, zero real) roots. $\endgroup$
    – JavaMan
    Feb 24, 2012 at 17:31
  • $\begingroup$ One other alternative: If you can find the vertex of the polynomial, then knowing the $y$-value of the vertex and knowing whether the quadratic function opens up or down will tell you how many times the quadratic crosses the $x$-axis. $\endgroup$
    – JavaMan
    Feb 24, 2012 at 17:33
  • $\begingroup$ Note that the quadratic equation has two roots -- both real roots or no real roots. $\endgroup$
    – user21436
    Feb 24, 2012 at 17:55
  • $\begingroup$ Just to nitpick: roots belong to equations, zeros belong to functions. $\endgroup$ Dec 23, 2018 at 11:23

2 Answers 2


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 $\mathbb i =\sqrt{-1})$.

  • The expression under the square root sign, if positive, we have two zeroes that are real (more commonly known as two distinct and Real roots).
  • If the expression under the square root sign is negative, we will get two zeroes that are Complex Numbers (more commonly known as two distinct but Complex roots).
  • You might ask what happens if the expression under the square root sign equals $0$? In that case, our relation would reduce down to $$x=\frac{-B}{2\cdot A} \pm 0$$ OR, $$x=\frac{-B}{2\cdot A}$$ So if the expression under the square root sign equals zero then you'd still get Real values for the zeroes but you'd get just a single "zero" for the equation (more commonly known as repeated and Real roots).

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?

  • If the roots are distinct and real, or, if the discriminant (btw, that is the name for the expression under the square root) is positive, then, the function should cross the x axis twice.Or 2 distinct real roots
  • Similarly, if the roots are Real and repeated, (i.e the discriminant is zero), the function should have only one point crossing (or in this case touching) the x-axis. Or repeated roots
  • Now if the roots are not real, then what? Because there are two distinct roots (but not real), it might be tempting to think that it also crosses the x- axis twice but, good luck finding the point where it actually crosses the x-axis! (where will you find a number $2+3\mathbb i $ on the x axis?). So the next logical deduction could be that it doesn't actually cross the x-axis at all, which is very plausible. And this is exactly what happens. OR, no real roots
    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

  • $\begingroup$ While I mentioned that $A$,$B$ and $C$ must always be real numbers,The OP should know that there can be quadratic functions with complex coefficients but, it is something that you need not bother right now (it is also irrelevant to your question) $\endgroup$
    – user11470
    Mar 26, 2012 at 5:46
  • $\begingroup$ Ridiculously thorough answer! Nice writing. $\endgroup$ Mar 31, 2012 at 6:01
  • $\begingroup$ Thank you. It means a lot :) $\endgroup$
    – user11470
    Mar 31, 2012 at 15:40

Depending on the level that you are at, you will find that most quadratics you are given will probably either factorise or have observable roots.

There is a measure of the number of real roots a quadratic has called the discriminant. This essentially comes from completing the square or looking at the quadratic formula.

If your quadratic is $ax^2 + bx + c$ then calculate $b^2 - 4ac$. If this is positive then you have $2$ different real roots. If it is zero then you have one (repeated) real root and. If it is negative then you have no real roots (they are complex numbers).


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