# Finding Roots of Quadratic (Equation, Examples & Graphs)

The root of a polynomial p(x) is the value of x, such that p(x) = 0. So finding the root is equivalent to solving the equation p(x) = 0. Usually, finding the roots of a higher degree polynomial is difficult. Fortunately, for a quadratic, we have a simple formula.

## Formula for Quadratic Roots (or Solutions)

The roots (or solutions) of a quadratic polynomial (or equation): Ax2 + Bx + C = 0 are given by the formula:

x1 = ( − B + √D)/(2A)

x2 = ( − B − √D)/(2A)

x1 and x2 are two roots of the quadratic. They also solve the equation.

A, B, and C are real numbers and A ≠ 0.

### Root Types & Number

A quadratic may have zero, one or two real roots. It depends on the value of discriminant D = B2 − 4AC:

Discriminant Root Type Number of Roots
D < 0 No Real Root 0
D = 0 Real 1
D > 0 Real 2

If we factorize the quadratic with single root x1, the factors have the form A(x − x1)(x − x1). Therefore x1 is also called a repeated root.

In the quadratic formula, we have a √D term. As the square root of a negative number cannot be real, for D < 0 no real root exists. However, we do have two complex roots.

## Steps for Finding Roots

### Step 1: Calculate Discriminant

Given the equation Ax2 + Bx + C = 0, calculate discriminant D = B2 − 4AC

### Step 2: Find Roots

If D = 0, the √D part of the formula vanishes. So there is only one (repeated) root:

x1 = x1 = ( − B)/(2A)

For other cases, the roots are:

x1 = ( − B + √D)/(2A)

x2 = ( − B − √D)/(2A)

Let us see an example.

### Example 1

Find roots of x2 − x − 12 = 0

#### Step 1: Calculate Discriminant

Coefficients: A = 1, B = − 1, C = − 12

We calculate the discriminant D = B2 − 4AC:

= ( − 1)2 − (4)(1)( − 12) = 1 + 48 = 49

#### Step 2: Find Roots

As D > 0, we get two real roots. Let’s find them.

x1 = ( − B + √D)/(2A)

= 8/2 = 4

The other root:

x2 = ( − B − √D)/(2A)

= − 6/2 = − 3

The two roots are 4 and −3.

Let’s see examples of the three cases and corresponding graphs.

## Root Type Examples

Following examples calculate the three types of roots. They relate them with the graph of the quadratic function.

### Example 2 – One Real Root

Find roots of the equation:

x2 + 6x + 9 = 0

#### Step 1: Calculate Discriminant

Coefficients: A = 1, B = 6, C = 9.

D = B2 − 4AC = 62 − (4 × 1 × 9)

= 36 − (36) = 0

For D = 0, we have only one root (solution).

#### Step 2: Find Roots

x1 = x2 = ( − B + √D)/(2A) = ( − B)/(2A)

= − 6/2 = − 3.

You can verify that x = − 3 indeed satisfies the equation.

If we plot the graph of quadratic function f(x) = x2 + 6x + 9, you can see that it attains the zero value at only one point, x = − 3.

### Example 3 – Two Real Roots

Solve the equation:

x2 − 3x + 2 = 0

#### Step 1: Calculate Discriminant

Coefficients: A = 1, B = − 3, C = 2.

The discriminant D = B2 − 4AC = ( − 3)2 − 4(1)(2)

= 9 − (8) = 1

#### Step 2: Find Roots

As D > 0, we have two roots.

x1 = ( − B + √D)/(2A)

= (3 + √1)/2

= 4/2 = 2

x2 = ( − B − √D)/(2A)

= (3 − √1)/2

= 2/2 = 1

Look at the graph of x2 − 3x + 2. It intersects the x-axis at both roots.

### Example 4 – No Real Root / Complex Roots

Solve −3x2 + 2x − 1 = 0

#### Step 1: Calculate Discriminant

Coefficients: A = -3, B = 2, C = -1.

Let us calculate discriminant D:

D = B2 − 4AC = (2)2 − 4( − 3)( − 1)

= 4 − (12) = − 9

#### Step 2: Find Roots

As D < 0, is not a real number. In this case there is no real solution for the equation.

The graph of −3x2 + 2x − 1 does not touch x-axis. Was this expected?

#### Calculate Imaginary/Complex Roots

If you have studied complex numbers, you can calculate the imaginary roots of the equation above.

x1 = ( − B + √D)/(2A)

= 1/3 − i/2

x2 = ( − B − √D)/(2A)

= 1/3 + i/2

## What are the Roots of Any Function?

For any any function f(x), its roots are values of x, such that f(x) = 0

In other words, root is a solution of the equation f(x) = 0. It is the value of x where the graph of f(x) intersects the x-axis.

In the following graph, f(x) has three roots: