# Math: How to Find the Roots of a Quadratic Function

*I studied applied mathematics, in which I did both a bachelor's and a master's degree.*

## Quadratic Functions

A quadratic function is a polynomial of degree two. That means it is of the form ax^2 + bx +c. Here, a, b and c can be any number. When you draw a quadratic function, you get a parabola as you can see in the picture above. When a is negative, this parabola will be upside down.

### What Are Roots?

The roots of a function are the points on which the value of the function is equal to zero. These correspond to the points where the graph crosses the x-axis. So when you want to find the roots of a function you have to set the function equal to zero. For a simple linear function, this is very easy. For example:

f(x) = x +3

Then the root is x = -3, since -3 + 3 = 0. Linear functions only have one root. Quadratic functions may have zero, one or two roots. An easy example is the following:

f(x) = x^2 - 1

When setting x^2-1 = 0, we see that x^2 = 1. This is the case for both x = 1 and x = -1.

An example of a quadratic function with only one root is the function x^2. This is only equal to zero when x is equal to zero. It might also happen that here are no roots. This is, for example, the case for the function x^2+3. Then, to find the root we have to have an x for which x^2 = -3. This is not possible, unless you use complex numbers. In most practical situations, the use of complex numbers does make sense, so we say there is no solution.

## Ways to Find the Roots of a Quadratic Function

### Factorization

The most common way people learn how to determine the the roots of a quadratic function is by factorizing. For a lot of quadratic functions this is the easiest way, but it also might be very difficult to see what to do. We have a quadratic function ax^2 + bx + c, but since we are going to set it equal to zero, we can divide all terms by a if a is not equal to zero. Then we have an equation of the form:

x^2 + px + q = 0.

Now we try to find factors s and t such that:

(x-s)(x-t) = x^2 + px + q

If we succeed we know that x^2 + px + q = 0 is true if and only if (x-s)(x-t) = 0 is true. (x-s)(x-t) = 0 means that either (x-s) = 0 or (x-t)=0. This means that x = s and x = t are both solutions, and hence they are the roots.

If (x-s)(x-t) = x^2 + px + q, then it holds that s*t = q and - s - t = p.

**Numerical Example**

x^2 + 8x + 15

Then we have to find s and t such that s*t = 15 and - s - t = 8. So if we choose s = -3 and t = -5 we get:

x^2 + 8x + 15 = (x+3)(x+5) = 0.

Hence, x = -3 or x = -5. Let's check these values: (-3)^2 +8*-3 +15 = 9 - 24 + 15 = 0 and (-5)^2 + 8*-5 +15 = 25 - 40 + 15 = 0. So indeed these are the roots.

It might however be very difficult to find such a factorization. For example:

x^2 -6x + 7

Then the roots are 3 - sqrt 2 and 3 + sqrt 2. These are not so easy to find.

**The ABC Formula**

Another way to find the roots of a quadratic function. This is an easy method that anyone can use. It is just a formula you can fill in that gives you roots. The formula is as follows for a quadratic function ax^2 + bx + c:

(-b + sqrt(b^2 -4ac))/2a and (-b - sqrt(b^2 -4ac))/2a

This formulas give both roots. When only one root exists both formulas will give the same answer. If no roots exist, then b^2 -4ac will be smaller than zero. Therefore the square root does not exist and there is no answer to the formula. The number b^2 -4ac is called the discriminant.

**Numeric example**

Let's try the formula on the same function we used for the example on factorizing:

x^2 + 8x + 15

Then a = 1, b = 8 and c = 15. Therefore:

(-b + sqrt(b^2 -4ac))/2a = (-8+sqrt(64-4*1*15))/2*1 = (-8+sqrt(4))/2 = -6/2 = -3

(-b - sqrt(b^2 -4ac))/2a = (-8-sqrt(64-4*1*15))/2*1 = (-8-sqrt(4))/2 = -10/2 = -5

So indeed, the formula gives the same roots.

### Completing the Square

The ABC Formula is made by using the completing the square method. The idea of completing the square is as follows. We have ax^2 + bx + c. We assume a = 1. If this would not be the case, we could divide by a and we get new values for b and c. The other side of the equation is zero, so if we divide that by a, it stays zero. Then we do the following:

x^2 + bx + c = (x+b/2)^2 -(b^2/4) + c = 0.

Then (x+b/2)^2 = (b^2/4) - c.

Therefore x+b/2 = sqrt((b^2/4) - c) or x+b/2 = - sqrt((b^2/4) - c).

This implies x = b/2+sqrt((b^2/4) - c) or x = b/2 - sqrt((b^2/4) - c).

This is equal to the ABC-Formula for a = 1. However, this is easier to calculate.

**Numerical Example**

We take again x^2 + 8x + 15. Then:

x^2 + 8x + 15 = (x+4)^2 -16+15 = (x+4)^2 -1 = 0.

Then x = -4 + sqrt 1 = -3 or x = -4 - sqrt 1 = -5.

So indeed, this gives the same solution as the other methods.

## Summary

We have seen three different methods to find the roots of a quadratic function of the form ax^2 + bx + c. The first was factorizing where we try to write the function as (x-s)(x-t). Then we know the solutions are s and t. The second method we saw was the ABC Formula. Here you just have to fill in a, b and c to get the solutions. Lastly, we had the completing the squares method where we try to write the function as (x-p)^2 + q.

## Higher Degree Functions

Determining the roots of a function of a degree higher than two is a more difficult task. For third-degree functions—functions of the form ax^3+bx^2+cx+d—there is a formula, just like the ABC Formula. This formula is pretty long and not so easy to use. For functions of degree four and higher, there is a proof that such a formula doesn't exist.

This means that finding the roots of a function of degree three is doable, but not easy by hand. For functions of degree four and higher, it becomes very difficult and therefore it can better be done by a computer.

## Comments

**Eyssant** on April 25, 2020:

A nice article.