Python Program to Find Roots of Quadratic Equation

Use the quadratic formula in Python: calculate discriminant with D = b**2 - 4*a*c, then find roots with root1 = (-b + D**0.5) / (2*a) and root2 = (-b - D**0.5) / (2*a).

Examples

Inputa=1, b=-3, c=2

OutputRoots are 2.0 and 1.0

Inputa=1, b=2, c=1

OutputRoots are -1.0 and -1.0

Inputa=1, b=0, c=1

OutputRoots are complex: 0.0 + 1.0j and 0.0 - 1.0j

How to Think About It

To find roots of a quadratic equation ax² + bx + c = 0, first calculate the discriminant D = b² - 4ac. If D is positive, roots are two real numbers; if zero, roots are one real number repeated; if negative, roots are complex numbers. Use the formula (-b ± √D) / (2a) to find the roots.

Algorithm

1

Get coefficients a, b, and c from the user

2

Calculate the discriminant D = b² - 4ac

3

If D is positive or zero, calculate two real roots using (-b ± √D) / (2a)

4

If D is negative, calculate complex roots using complex square root

5

Print the roots

Code

python

a = float(input('Enter coefficient a: '))
b = float(input('Enter coefficient b: '))
c = float(input('Enter coefficient c: '))

if a == 0:
    print('Coefficient a cannot be zero for a quadratic equation.')
else:
    d = b**2 - 4*a*c

    if d >= 0:
        root1 = (-b + d**0.5) / (2*a)
        root2 = (-b - d**0.5) / (2*a)
        print(f'Roots are {root1} and {root2}')
    else:
        real_part = -b / (2*a)
        imag_part = (abs(d)**0.5) / (2*a)
        print(f'Roots are complex: {real_part} + {imag_part}j and {real_part} - {imag_part}j')

Output

Enter coefficient a: 1
Enter coefficient b: -3
Enter coefficient c: 2
Roots are 2.0 and 1.0

Dry Run

Let's trace the example a=1, b=-3, c=2 through the code

1

Input coefficients

a=1, b=-3, c=2

2

Calculate discriminant

d = (-3)**2 - 4*1*2 = 9 - 8 = 1

3

Check discriminant sign

d = 1 >= 0, so roots are real

4

Calculate roots

root1 = (-(-3) + 1**0.5) / (2*1) = (3 + 1) / 2 = 2.0 root2 = (-(-3) - 1**0.5) / (2*1) = (3 - 1) / 2 = 1.0

5

Print roots

Roots are 2.0 and 1.0

Step	Discriminant (d)	Root1	Root2
Calculate d	1
Calculate root1	1	2.0
Calculate root2	1	2.0	1.0

Why This Works

Step 1: Calculate discriminant

The discriminant D = b**2 - 4*a*c tells us the nature of roots: positive means two real roots, zero means one real root, negative means complex roots.

Step 2: Compute roots for real discriminant

When D >= 0, roots are calculated using (-b ± sqrt(D)) / (2*a) which gives two real numbers.

Step 3: Handle complex roots

If D < 0, roots are complex. We calculate real and imaginary parts separately and display them as complex numbers.

Alternative Approaches

Using cmath module for all cases

python

import cmath

a = float(input('Enter a: '))
b = float(input('Enter b: '))
c = float(input('Enter c: '))

if a == 0:
    print('Coefficient a cannot be zero for a quadratic equation.')
else:
    d = cmath.sqrt(b**2 - 4*a*c)
    root1 = (-b + d) / (2*a)
    root2 = (-b - d) / (2*a)
    print(f'Roots are {root1} and {root2}')

This method handles real and complex roots uniformly but uses the cmath module which always returns complex numbers.

Using numpy.roots function

python

import numpy as np
coeffs = [1, -3, 2]
roots = np.roots(coeffs)
print(f'Roots are {roots[0]} and {roots[1]}')

This method is concise and uses a library function but requires numpy installed and does not show step-by-step calculation.

Complexity: O(1) time, O(1) space

Time Complexity

The program performs a fixed number of arithmetic operations and conditional checks, so it runs in constant time.

Space Complexity

Only a few variables are used to store inputs and results, so space usage is constant.

Which Approach is Fastest?

All approaches run in constant time; using cmath simplifies code for complex roots but adds module overhead, while numpy is heavier but useful for polynomial roots in general.

Approach	Time	Space	Best For
Manual calculation with if-else	O(1)	O(1)	Learning and simple scripts
Using cmath module	O(1)	O(1)	Handling complex roots easily
Using numpy.roots	O(1)	O(1)	Polynomials of any degree with library support