9. Python | 自定义函数_求一元二次方程根(代码)

练习

请定义一个函数quadratic(a, b, c)，接收3个参数，返回一元二次方程：ax2 + bx + c = 0的两个解。
提示：计算平方根可以调用math.sqrt()函数：

# -*- coding:utf-8 -*-
import math 
def quadratic(a,b,c):
    if not isinstance(a+b+c,(int,float)):
        raise TypeError('bad operand type')
    elif a != 0:
        # ∆ = b²-4ac > 0时,有两个解   
        if b*b - 4*a*c > 0:
            x1 = (-b + math.sqrt(b*b - 4*a*c))/2/a
            x2 = (-b - math.sqrt(b*b - 4*a*c))/2/a
            return ('x1 =',x1,'x2 =',x2) 
        # Δ = b²-4ac = 0时,有一个解
        elif b*b - 4*a*c == 0:
            return ('x =',-b/2/a)
        # Δ = b²-4ac < 0时,无解
        else:
            return ('x无解')
    else:   # a = 0时,变为一元一次方程
        if b != 0:
            return ('x =',-c/b)
        else:
            if c == 0:  # b = 0,c = 0时,x可以取任意值
                return ('x为任意值')
            else:       # b = 0,c != 0时,方程无解
                return ('x无解')
# 测试:
'''
print('quadratic(2, 3, 1) =', quadratic(2, 3, 1))
print('quadratic(1, 3, -4) =', quadratic(1, 3, -4))

if quadratic(2, 3, 1) != (-0.5, -1.0):
    print('测试失败')
elif quadratic(1, 3, -4) != (1.0, -4.0):
    print('测试失败')
else:
    print('测试成功')'''
   
print('quadratic(0,3,3)=',quadratic(0,3,3))