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Update quadratic equations solver (#1764)

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Use pythons complex number module cmath for the calculation of the
roots
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TheSuperNoob 2020-02-19 19:45:55 +01:00 committed by GitHub
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maths/quadratic_equations_complex_numbers.py
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@ -1,38 +1,36 @@
from math import sqrt
from cmath import sqrt
from typing import Tuple
def QuadraticEquation(a: int, b: int, c: int) -> Tuple[str, str]:
def quadratic_roots(a: int, b: int, c: int) -> Tuple[complex, complex]:
"""
Given the numerical coefficients a, b and c,
prints the solutions for a quadratic equation, for a*x*x + b*x + c.
calculates the roots for any quadratic equation of the form ax^2 + bx + c
>>> QuadraticEquation(a=1, b=3, c=-4)
('1.0', '-4.0')
>>> QuadraticEquation(5, 6, 1)
('-0.2', '-1.0')
>>> quadratic_roots(a=1, b=3, c=-4)
(1.0, -4.0)
>>> quadratic_roots(5, 6, 1)
(-0.2, -1.0)
>>> quadratic_roots(1, -6, 25)
((3+4j), (3-4j))
"""
if a == 0:
raise ValueError("Coefficient 'a' must not be zero for quadratic equations.")
raise ValueError("Coefficient 'a' must not be zero.")
delta = b * b - 4 * a * c
if delta >= 0:
return str((-b + sqrt(delta)) / (2 * a)), str((-b - sqrt(delta)) / (2 * a))
"""
Treats cases of Complexes Solutions(i = imaginary unit)
Ex.: a = 5, b = 2, c = 1
Solution1 = (- 2 + 4.0 *i)/2 and Solution2 = (- 2 + 4.0 *i)/ 10
"""
snd = sqrt(-delta)
if b == 0:
return f"({snd} * i) / 2", f"({snd} * i) / {2 * a}"
b = -abs(b)
return f"({b}+{snd} * i) / 2", f"({b}+{snd} * i) / {2 * a}"
root_1 = (-b + sqrt(delta)) / (2 * a)
root_2 = (-b - sqrt(delta)) / (2 * a)
return (
root_1.real if not root_1.imag else root_1,
root_2.real if not root_2.imag else root_2,
)
def main():
solutions = QuadraticEquation(a=5, b=6, c=1)
print("The equation solutions are: {} and {}".format(*solutions))
# The equation solutions are: -0.2 and -1.0
solutions = quadratic_roots(a=5, b=6, c=1)
print("The solutions are: {} and {}".format(*solutions))
if __name__ == "__main__":