python - 返回错误的正#

Question

我想要做的是写一个二次方程求解器,但是当解决方案应该是-1,就像quadratic(2, 4, 2)它返回一样1

我究竟做错了什么？

#!/usr/bin/python
import math
def quadratic(a, b, c):
        #a = raw_input("What\'s your `a` value?\t")
        #b = raw_input("What\'s your `b` value?\t")
        #c = raw_input("What\'s your `c` value?\t")
        a, b, c = float(a), float(b), float(c)
        disc = (b*b)-(4*a*c)
        print "Discriminant is:\n" + str(disc)
        if disc >= 0:
                root = math.sqrt(disc)
                top1 = b + root
                top2 = b - root
                sol1 = top1/(2*a)
                sol2 = top2/(2*a)
                if sol1 != sol2:
                        print "Solution 1:\n" + str(sol1) + "\nSolution 2:\n" + str(sol2)
                if sol1 == sol2:
                        print "One solution:\n" + str(sol1)
        else:
                print "No solution!"

编辑:它返回以下内容......

>>> import mathmodules
>>> mathmodules.quadratic(2, 4, 2)
Discriminant is:
0.0
One solution:
1.0

Answer 1

除非我上学以后公式发生了变化(人们永远不能确定),所以(-b +- sqrt(b^2-4ac)) / 2a,你b的代码中有.

[编辑]我可以建议一个重构器吗？

def quadratic(a, b, c):
    discriminant = b**2 - 4*a*c
    if discriminant < 0:
      return []
    elif discriminant == 0:
      return [-b / (2*a)]
    else:
      root = math.sqrt(discriminant)
      return [(-b + root) / (2*a), (-b - root) / (2*a)]

print quadratic(2, 3, 2) # []
print quadratic(2, 4, 2) # [-1]                    
print quadratic(2, 5, 2) # [-0.5, -2.0]