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:
Related
Quadratic Function Graph (Illustrations) ➤