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二次方程求解器不工作
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Stack Overflow用户
提问于 2015-05-17 14:57:32
回答 4查看 287关注 0票数 1

我正在尝试创建一个二次方程求解器,但当我输入一个大于1的系数时,它似乎不起作用?代码和错误消息如下。任何帮助都是非常感谢的。

代码语言:javascript
复制
print "Welcome to the quadratic equation solver."
print "The general quadratic equation = ax^2 + bx + c.\n"

def POS(a,b):
#This function gives the point at which the quadratic turns
    tp = float((-b)/(a*2))
    return (tp)

#This allows for the user to input the values of the variables
while True:
    s = raw_input ("Please insert a numerical value for a: ")
    try:
        a = float(s)
        break
    except ValueError:
        print "Please enter a numeric value."
print ("Well done.\n")

while True:
    s = raw_input ("Please insert a numerical value for b: ")
    try:
        b = float(s)
        break
    except ValueError:
        print "Please enter a numeric value."
print ("Well done.\n")

while True:
    s = raw_input ("Please insert a numerical value for c: ")
    try:
        c = float(s)
        break
    except ValueError:
        print "Please enter a numeric value."
print ("Well done.\n")

#This uses the function to give the co-ordinate of the turning point
print POS(a,b), "is the x-value of the turning point"
print ((a)*(POS(a,b)**2))+((b)*POS(a,b))+c, "is they y-value of the turning point. \n"

#This tells whether the quadratic is positive or negative
if a >0:
    print "The quadratic is positive.\n"

if a<0:
    print "The quadratic is negative.\n"

#This determines the root of the quadratic
root1 = (-b +((b**2) - (4*a*c))**0.5) / (2 * a)
root2 = (-b -((b**2) - (4*a*c))**0.5) / (2 * a)
print "The quadratic has a root at x =",root1
if root1 != root2:
    print "The quadratic has another root at x =",

#This uses the discriminant to determine the nature of the quadratic
if (b**2)-(4*a*c) == 0:
    print root1

elif (b**2)-(4*a*c) > 0:
    print root1 and root2

elif (b**2)-(4*a*c) < 0:
    print "The quadratic contains no roots"

for x in range(-1000,1000):
    quadratic = (a*x**2)+(b*x)+c

#Determines the derivitive and second derivitive of the quadratic
print "The derivitive of the quadratic is: ",2*a,"x", "+",b
print "The second derivitive of the quadratic is"

#Create plot
X = arange (-1000,1000,1)
plot(X, quadratic, linewidth=3, label="quadratic")

#Create title and axes label
title ("Graph of the quadratic")
xlabel ("x")
ylabel ("y")

#grid
grid(True)
legend()
show()

错误消息:

代码语言:javascript
复制
Traceback (most recent call last):
  File "C:\Users\Peter\Downloads\test.py", line 49, in <module>
    root1 = (-b +((b**2) - (4*a*c))**0.5) / (2 * a)
ValueError: negative number cannot be raised to a fractional power
EN

回答 4

Stack Overflow用户

发布于 2015-05-17 15:41:52

代码的一部分:

代码语言:javascript
复制
#This determines the root of the quadratic
root1 = (-b +((b**2) - (4*a*c))**0.5) / (2 * a)
root2 = (-b -((b**2) - (4*a*c))**0.5) / (2 * a)
print "The quadratic has a root at x =",root1
if root1 != root2:
    print "The quadratic has another root at x =",

#This uses the discriminant to determine the nature of the quadratic
if (b**2)-(4*a*c) == 0:
    print root1

elif (b**2)-(4*a*c) > 0:
    print root1 and root2

elif (b**2)-(4*a*c) < 0:
    print "The quadratic contains no roots"

for x in range(-1000,1000):
    quadratic = (a*x**2)+(b*x)+c

让我们用D替换(b**2)-(4*a*c)

代码语言:javascript
复制
#This determines the root of the quadratic
D = (b**2)-(4*a*c)

root1 = (-b +(D)**0.5) / (2 * a)
root2 = (-b -(D)**0.5) / (2 * a)

print "The quadratic has a root at x =", root1
if root1 != root2:
    print "The quadratic has another root at x =",

#This uses the discriminant to determine the nature of the quadratic
if D == 0:
    print root1

elif D > 0:
    # old code:
    # print root1 and root2
    # new code:
    print root1, root2

elif D < 0:
    print "The quadratic contains no roots"

for x in range(-1000,1000):
    quadratic = (a*x**2)+(b*x)+c

这两行中的问题:

代码语言:javascript
复制
root1 = (-b +(D)**0.5) / (2 * a)
root2 = (-b -(D)**0.5) / (2 * a)

如果D小于0,则此行将引发您获得的ValueError。作为错误消息,所述数字不能被提高到分数幂。所以我们必须检查D是否小于0。

代码语言:javascript
复制
#This determines the root of the quadratic
D = (b**2)-(4*a*c)

# new code:
if D >= 0:
    root1 = (-b +(D)**0.5) / (2 * a)
    root2 = (-b -(D)**0.5) / (2 * a)

    print "The quadratic has a root at x =", root1
    if root1 != root2:
        print "The quadratic has another root at x =", root2


#This uses the discriminant to determine the nature of the quadratic
# We've already printed root1 and root2
# if D == 0:
#     print root1

# elif D > 0:
#     print root1, root2

# D < 0
else:
    print "The quadratic contains no roots"

for x in range(-1000,1000):
    quadratic = (a*x**2)+(b*x)+c
票数 2
EN

Stack Overflow用户

发布于 2015-05-17 15:56:45

您还可以切换到Python3,这样您的号码就不再是ValueError,而是implicitly cast to a complex number.

代码语言:javascript
复制
(-2)**.2
(0.9293164906031477+0.6751879523998812j)
票数 2
EN

Stack Overflow用户

发布于 2015-05-17 15:29:52

代码语言:javascript
复制
#This uses the discriminant to determine the nature of the quadratic
if (b**2)-(4*a*c) == 0:
    print root1

elif (b**2)-(4*a*c) > 0:
    print root1 and root2

elif (b**2)-(4*a*c) < 0:
    print "The quadratic contains no roots"

在处理根之前执行此检查,如果是复杂根,则中断。

代码语言:javascript
复制
>>> a=2
>>> b=4
>>> c=1
>>> delta=math.pow(b,2)-4*a*c
>>> math.sqrt(delta) #raises error for -ve integers
3.4641016151377544 
gamma=math.sqrt(delta)
>>>root1=(-b+gamma)/2/a
>>>root2=(-b-gamma)/2/a
票数 1
EN
页面原文内容由Stack Overflow提供。腾讯云小微IT领域专用引擎提供翻译支持
原文链接:

https://stackoverflow.com/questions/30284231

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