用python求一元二次方程的根，用“行”if语句和if语句的嵌套两种方法分别求解

在Python中，求解一元二次方程 \(ax^2 + bx + c = 0\) 的根，可以利用公式法，也可以通过数值计算库如`numpy`或者直接编写一些条件判断。以下是使用if语句和if语句的嵌套两种方法：

**方法一：if语句**

import cmath

def quadratic_solver(a, b, c):
    delta = (b ** 2) - (4 * a * c)
    
    if delta > 0:
        root1 = (- cmath.sqrt(delta)) / (2 * a)
        print(f"两个实数根: {root1.real:.2f}, {root2.real:.2f}")
    elif delta == 0:
        root = -b / (2 * a)
        print(f"唯一实数根: {root:.2f}")
    else:  # 当delta < 0，有两个复数根
        real_part = -b / (2 * a)
        imaginary_part = cmath.sqrt(-delta) / (2 * a)
        print(f"两个复数根: {real_part:.2f} + {imaginary_part:.2f}i, {real_part:.2f} - {imaginary_part:.2f}i")

# 调用函数并传入系数
quadratic_solver(1, -3, 2)

**方法二：if语句的嵌套**

def quadratic_solver_nested(a, b, c):
    discriminant = b ** 2 - 4 * a * c
    
    if discriminant >= 0:
        if discriminant == 0:
            root = -b / (2 * a)
            print("唯一实数根:", root)
        else:
            sqrt_discriminant = discriminant ** 0.5
            root1 = (-b + sqrt_discriminant) / (2 * a)
            root2 = (-b - sqrt_discriminant) / (2 * a)
            print("两个实数根:", root1, root2)
    else:
        complex_part = discriminant ** 0.5
        print("两个复数根:", (-b + complex_part * 1j) / (2 * a), "and", (-b - complex_part * 1j) / (2 * a))

# 使用同样的系数调用函数
quadratic_solver_nested(1, -3, 2)