Python program to solve quadratic equation
Given a quadratic equation the task is solve the equation or find out the roots of the equation. Standard form of quadratic equation is –
ax2 + bx + c where, a, b, and c are coefficient and real numbers and also a ≠ 0. If a is equal to 0 that equation is not valid quadratic equation.
Examples:
Input :a = 1, b = 2, c = 1 Output : Roots are real and same -1.0 Input :a = 2, b = 2, c = 1 Output : Roots are complex -0.5 + i 2.0 -0.5 - i 2.0 Input :a = 1, b = 10, c = -24 Output : Roots are real and different 2.0 -12.0
Method 1: Using the direct formula
Using the below quadratic formula we can find the root of the quadratic equation.
There are following important cases.
If b*b < 4*a*c, then roots are complex (not real). For example roots of x2 + x + 1, roots are -0.5 + i1.73205 and -0.5 - i1.73205 If b*b == 4*a*c, then roots are real and both roots are same. For example, roots of x2 - 2x + 1 are 1 and 1 If b*b > 4*a*c, then roots are real and different. For example, roots of x2 - 7x - 12 are 3 and 4
# Python program to find roots of quadratic equationimport math # function for finding rootsdef equationroots( a, b, c): # calculating discriminant using formula dis = b * b - 4 * a * c sqrt_val = math.sqrt(abs(dis)) # checking condition for discriminant if dis > 0: print(" real and different roots ") print((-b + sqrt_val)/(2 * a)) print((-b - sqrt_val)/(2 * a)) elif dis == 0: print(" real and same roots") print(-b / (2 * a)) # when discriminant is less than 0 else: print("Complex Roots") print(- b / (2 * a), " + i", sqrt_val) print(- b / (2 * a), " - i", sqrt_val) # Driver Program a = 1b = 10c = -24 # If a is 0, then incorrect equationif a == 0: print("Input correct quadratic equation") else: equationroots(a, b, c) |
Output:
real and different roots 2.0 -12.0
Method 2: Using the complex math module
First, we have to calculate the discriminant and then find two solution of quadratic equation using cmath module.
# import complex math moduleimport cmath a = 1b = 4c = 2 # calculating the discriminantdis = (b**2) - (4 * a*c) # find two resultsans1 = (-b-cmath.sqrt(dis))/(2 * a)ans2 = (-b + cmath.sqrt(dis))/(2 * a) # printing the resultsprint('The roots are')print(ans1)print(ans2) |
Output:
The roots are (-3.414213562373095+0j) (-0.5857864376269049+0j)
Last Updated on November 13, 2021 by admin