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Python | Finding Solutions of a Polynomial Equation
  • Difficulty Level : Basic
  • Last Updated : 27 Dec, 2019

Given a quadratic equation, the task is to find the possible solutions to it.

Examples:

Input : 
enter the coef of x2 : 1
enter the coef of x  : 2
enter the costant    : 1
Output :
the value for x is -1.0

Input :
enter the coef of x2 : 2
enter the coef of x  : 3
enter the costant    : 2
Output :
x1 = -3+5.656854249492381i/4 and x2 = -3-5.656854249492381i/4

Algorithm :

Start.
Prompt the values for a, b, c. 
Compute i = b**2-4*a*c
If i get negative value g=square root(-i)
Else h = sqrt(i)
Compute e = -b+h/(2*a)
Compute f = -b-h/(2*a)
If condition e==f then
    Print e
Else
    Print e and f
If i is negative then
    Print -b+g/(2*a) and -b-g/(2*a)
stop

Below is the Python implementation of the above mentioned task.




# Python program for solving a quadratic equation.
   
from math import sqrt   
try:     
  
    # if user gives non int values it will go to except block
    a = 1
    b = 2
    c = 1
    i = b**2-4 * a * c
  
    # magic condition for complex values
    g = sqrt(-i)
    try:
        d = sqrt(i)
        # two resultants
        e = (-b + d) / 2 * a  
        f = (-b-d) / 2 * a
        if e == f:
            print("the values for x is " + str(e))
        else:
            print("the value for x1 is " + str(e) +
                  " and x2 is " + str(f))
    except ValueError:
        print("the result for your equation is in complex")
          
        # to print complex resultants.
        print("x1 = " + str(-b) + "+" + str(g) + "i/" + str(2 * a) + 
              " and x2 = " + str(-b) + "-" + str(g) + "i/" + 
              str(2 * a))   
          
except ValueError:
    print("enter a number not a string or char")

Output :



the values for x is -1.0

Explanation :
First, this program will get three inputs from the user. The values are the coefficient of x2, coefficient of x and constant. Then it performs the formula
(-b + (or) - sqrt(b2 - 4 * a * c) / 2a)
For complex the value of (b2 - 4 * a * c) gets negative. Rooting negative values will throw a value error. In this case, turn the result of -(b2 - 4 * a * c) and then root it. Don’t forget to include i at last.

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