Python 解一元二次方程ax^2+bx+c=0

from math import *a = float(input("a:"))b = float(input("b:"))c = float(input("c:"))if a == 0:print("x=",-c/b)else:beta = b**2-4*a*c #b**2是b的平方if beta > 0:x1 = (-b + sqrt(beta))/(2*a)#记得大括号x2 = (-b

淡小愉

14399人浏览 · 2022-03-03 21:13:21

淡小愉 · 2022-03-03 21:13:21 发布

from math import *

a = float(input("a:"))
b = float(input("b:"))
c = float(input("c:"))    
if a == 0:
    print("x=",-c/b)
else:
    beta = b**2-4*a*c #b**2是b的平方
    if beta > 0:
        x1 = (-b + sqrt(beta))/(2*a)  #记得大括号
        x2 = (-b - sqrt(beta))/(2*a)
        print("x1=",x1," x2=",x2)
    elif beta == 0:
        x = (-b + sqrt(beta))/(2*a)
        print("x=",x)
    else:
        real = -b/(2*a)
        imag = beta/(2*a)
        print("x1=",complex(real,imag),"x2=",complex(real,-imag))

测试结果：

a:1
b:1
c:5
x1= (-0.5-9.5j) x2= (-0.5+9.5j)