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Python Program to Solve a Quadratic Equation

Intermediate · Numbers

A quadratic equation has the form ax² + bx + c = 0, where a is not 0, and its roots come from the formula x = (-b ± √(b² - 4ac)) / 2a. The discriminant d = b² - 4ac decides the kind of roots: two real roots if d is positive, one repeated root if d is 0, and two complex roots if d is negative. This program solves three sample equations, one of each kind.

Program

quadratic_equation.py
import math
# Each sample is (a, b, c) for a*x^2 + b*x + c = 0, with a not 0
samples = [(1, -5, 6), (1, -4, 4), (1, 2, 5)]
# or: a = float(input("Enter a: ")), and the same for b and c
for a, b, c in samples:
d = b * b - 4 * a * c # the discriminant
print(f"a = {a}, b = {b}, c = {c}, discriminant = {d}")
if d > 0:
root1 = (-b + math.sqrt(d)) / (2 * a)
root2 = (-b - math.sqrt(d)) / (2 * a)
print(" Two real roots:", root1, "and", root2)
elif d == 0:
root = -b / (2 * a)
print(" One repeated real root:", root)
else:
real = -b / (2 * a)
imag = math.sqrt(-d) / (2 * abs(a))
print(f" Two complex roots: {real} + {imag}i and {real} - {imag}i")
Output
a = 1, b = -5, c = 6, discriminant = 1
  Two real roots: 3.0 and 2.0
a = 1, b = -4, c = 4, discriminant = 0
  One repeated real root: 2.0
a = 1, b = 2, c = 5, discriminant = -16
  Two complex roots: -1.0 + 2.0i and -1.0 - 2.0i

How it works

  1. The discriminant d = b * b - 4 * a * c is worked out first, because its sign decides which case applies.
  2. If d > 0, math.sqrt(d) is a real number and the plus and minus versions of the formula give two different roots.
  3. If d == 0, both versions give the same value, -b / (2 * a), so there is one repeated root.
  4. If d < 0, math.sqrt(d) would raise an error, so the program takes the square root of -d and prints it as the imaginary part, marked with i.

Another way to write it

Using cmath for every case

Using cmath for every case
import cmath
samples = [(1, -5, 6), (1, -4, 4), (1, 2, 5)]
for a, b, c in samples:
d = b * b - 4 * a * c
# cmath.sqrt also works on negative numbers, giving a complex result
root1 = (-b + cmath.sqrt(d)) / (2 * a)
root2 = (-b - cmath.sqrt(d)) / (2 * a)
print(f"a = {a}, b = {b}, c = {c}: roots {root1} and {root2}")
Output
a = 1, b = -5, c = 6: roots (3+0j) and (2+0j)
a = 1, b = -4, c = 4: roots (2+0j) and (2+0j)
a = 1, b = 2, c = 5: roots (-1+2j) and (-1-2j)

The cmath module can take the square root of a negative number, so one formula covers all three cases. Python writes the imaginary unit as j, and real roots appear with +0j.

Try it yourself

Add the equation 2x² + 3x - 2 = 0 to samples and check that its roots are 0.5 and -2.

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