Python program to solve quadratic equation

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robaa

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2022-11-06

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Quadratic equation:

Quadratic equation is made from a Latin term "quadrates" which means square. It is a special type of equation having the form of:

ax2+bx+c=0

Here, "x" is unknown which you have to find and "a", "b", "c" specifies the numbers such that "a" is not equal to 0. If a = 0 then the equation becomes liner not quadratic anymore.

In the equation, a, b and c are called coefficients

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robaa

2022-11-06

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# import complex math module  
import cmath  
a = float(input('Enter a: '))  
b = float(input('Enter b: '))  
c = float(input('Enter c: '))  
  
# calculate the discriminant  
d = (b**2) - (4*a*c)  
  
# find two solutions  
sol1 = (-b-cmath.sqrt(d))/(2*a)  
sol2 = (-b+cmath.sqrt(d))/(2*a)  
print('The solution are {0} and {1}'.format(sol1,sol2))

Output:

Enter a: 8
Enter b: 5
Enter c: 9
The solution are (-0.3125-1.0135796712641785j) and (-0.3125+1.0135796712641785j)

Explanation -

In the first line, we have imported the cmath module and we have defined three variables named a, b, and c which takes input from the user. Then, we calculated the discriminant using the formula. Using the cmath.sqrt() method, we have calculated two solutions and printed the result.

Second Method

We can get the solution of the quadric equation by using direct formula. Let's understand the following example.

Above formula consists of the following cases.

If b2 < 4ac, then the roots are complex (not real). For example - x2 + x + 1, roots are -0.5 + i1.73205 and +0.5 - i1.73205.
If b2 == 4ac, then the both roots are same For example - x2 + x + 1, roots are -0.5 + i1.73205 and +0.5 - i1.73205.
If b2 > 4ac, then the roots are real and different. For example - x2 - 7 x - 12, roots are 3 and 4.

Example -

# Python program to find roots of quadratic equation  
import math  
  
  
# function for finding roots  
def findRoots(a, b, c):  
  
    dis_form = b * b - 4 * a * c  
    sqrt_val = math.sqrt(abs(dis_form))  
  
  
    if dis_form > 0:  
        print(" real and different roots ")  
        print((-b + sqrt_val) / (2 * a))  
        print((-b - sqrt_val) / (2 * a))  
  
    elif dis_form == 0:  
        print(" real and same roots")  
        print(-b / (2 * a))  
  
  
    else:  
        print("Complex Roots")  
        print(- b / (2 * a), " + i", sqrt_val)  
        print(- b / (2 * a), " - i", sqrt_val)  
  
  
a = int(input('Enter a:'))  
b = int(input('Enter b:'))  
c = int(input('Enter c:'))  
  
# If a is 0, then incorrect equation  
if a == 0:  
    print("Input correct quadratic equation")  
  
else:  
    findRoots(a, b, c)

Output:

Enter a:7
Enter b:5
Enter c:2
Complex Roots
-0.35714285714285715  + i 5.5677643628300215
-0.35714285714285715  - i 5.5677643628300215

Explanation -

In the above code, we have imported the math module and defined the formula to calculate discriminant. We have then defined findRoots function which take three integer values as arguments. Then we have checked the roots using the if-elif-else statement.

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