0#求解一元二次方程ax²+bx+c=0的两个解。
Topic sourceimport mathdef quadratic(a, b, c): if not isinstance(a + b + c, (int, float)): raise TypeError('bad operand type') z = math.pow(b, 2) - 4 * a * c if z >= 0 and a != 0: x1 = (-b + math.sqrt(z)) / (2 * a) x2 = (-b - math.sqrt(z)) / (2 * a) return x1, x2 if z >= 0 and a == 0 and b == 0 and c == 0: print('x值为任意值') if z >= 0 and a == 0 and b != 0: x1 = -c / b x2 = x1 return x1, x2 if z < 0: print('方程无解')
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