Question

The roots (solutions) to the quadratic equation ax2 +bx+c = 0 is given by the well-known formula x = 2a −b ± √b −4ac 2

Write a python program that takes the three coefficients a, b, and c as input, computes and displays the solutions according to the aforementioned formula. Note the ± in the numerator means there are two answers. But this will not apply if a is zero, so that condition must be checked separately; design your program such that: 1. If a = 0 and b = 0 and c = 0, it prints “solution is indeterminate!”. 2. If a = 0 and b = 0 but c =/ 0 ,it prints “there is no solution”. 3. If a = 0 but b =/ 0 and c =/ 0 , it prints “equation is linear”, computes the solution as x = −c/b and display it. The above formula also fails to work (for real numbers) if the expression under the square root is negative. That expression b ac is called the discriminant of the 2 − 4 quadratic. Define that as a separate variable d and check its sign: 1. If d > 0 then there are two real roots given by the formula above. 2. If d = 0 then the roots are loaded and equal to −b/2a. 3. If d < 0 then there is no real solution (roots are complex numbers).

Consider the following inputs/outputs for your test cases. Input Output a b c 1 -1 -2 There are two real roots. x = 2 and -1 0 -1 -2 The equation is linear. x = -2 0 0 -2 There is no solution. 0 0 0 Solution is indeterminate! 2 4 2 Roots are loaded. x = -1 5 2 1 there is no real solution.

Answer 1

Code (Intedentation is important So I am giving screenshot also)

a=int(input("Enter a: ")) b=int(input("Enter b: ")) c=int(input("Enter c: ")) if(a==0):    if(b==0 and c==0):#1st condition        print("Solution Indeterminate")    elif(b==0 and c!=0):#2nd condition        print("No Solution")    elif(b!=0 and c!=0):# 3rd condition        print("Equation is linear")        x=(-c/b)        print("x = ",x)#Printing root for linear else:      d=((b*b)-(4*a*c)) # finding condition of roots    if(d<0):#Complex roots        print("No real solution")    elif(d==0):#Equal roots        print("Roots are loaded")        x=(-b/(2*a))        print("x = ",x)    else:#Real roots        print("Real Roots")        x1=(-b+(d)**0.5)/(2*a)        x2=   (-b-(d)**0.5)/(2*a)        print("x1 = ",x1)#printing root1        print("x2 = ",x2)#printing root2

Output:-