TypeError: 'int' object is not callable: Quadratic

Question

(Finding the values of x in a quadratic equation w/o importing.) Whenever I run the program, Python stops at discriminant = (b ** 2) - 4(a * c) and shows TypeError: 'int' object is not callable. What's wrong?

#------SquareRootDefinition---------#
def Square_Root(n, x):
if n > 0:
    y = (x + n/x) / 2
    while x != y:
        x = y
        return Square_Root(n, x)
    else:
        if abs(10 ** -7) > abs(n - x ** 2):
            return y
elif n == 0:
    return 0
else:
    return str(int(-n)) + "i"

#----------Quadratic Equation--------------#

a = input("Enter coefficient a: ")
while a == 0:
    print "a must not be equal to 0."
    a = input("Enter coefficient a: ")
b = input("Enter coefficient b: ")
c = input("Enter coefficient c: ")

def Quadratic(a, b, c):
    discriminant = (b ** 2) - 4(a * c)
    if discriminant < 0:
        print "imaginary"
    elif discriminant >= 0:
        Sqrt_Disc = Square_Root(discriminant)
        First_Root = (-b + Sqrt_Disc) / (2 * a)
        Second_Root = (-b - Sqrt_Disc) / (2 * a)

  return First_Root, Second_Root

X_1, X_2 = Quadratic(a, b, c)

BTW, the code's not finished yet. – user1947570 Jan 04 '13 at 05:54 — user1947570, Jan 04 '13 at 05:54

score 7 · Answer 1 · answered Jan 04 '13 at 05:54

7

4(a * c) is not valid Python. You mean 4 * a * c. You can use juxtaposition and omit the multiplication symbol in mathematical notation, but not in Python (or most other programming languages).

answered Jan 04 '13 at 05:54

BrenBarn

242,874
37
412
384

Thank you @BrenBarn. You mean the only issue is the operation? – user1947570 Jan 04 '13 at 05:56
@user1947570: The only issue is the way you wrote the operation. You have to actually write the multiplication symbol (`*`). – BrenBarn Jan 04 '13 at 06:01
Imagine if you wrote `a (b * c)`. Is `a` a function? Or is `a` something you want to multiply? Because of this ambiguity, multiplication cannot be implicit. – David Schwartz Jan 04 '13 at 06:07

Martijn Pieters · Accepted Answer · 2013-01-04T06:09:44.547

You are trying to use 4 as a function:

discriminant = (b ** 2) - 4(a * c)

You missed *:

discriminant = (b ** 2) - 4 * (a * c)

Also, if your discriminant turns out to be lower than 0, you'll get an unbound local exception:

>>> X_1, X_2 = Quadratic(2, 1, 1)
imaginary
Traceback (most recent call last):
  File "<stdin>", line 1, in <module>
  File "<stdin>", line 9, in Quadratic
UnboundLocalError: local variable 'First_Root' referenced before assignment

You'll need to add a return there, or better still, raise an exception:

def Quadratic(a, b, c):
    discriminant = (b ** 2) - 4(a * c)
    if discriminant < 0:
        raise ValueError("imaginary")
    elif discriminant >= 0:
        Sqrt_Disc = Square_Root(discriminant)
        First_Root = (-b + Sqrt_Disc) / (2 * a)
        Second_Root = (-b - Sqrt_Disc) / (2 * a)

    return First_Root, Second_Root

Your Square_Root() function is missing it's default value for x:

def Square_Root(n, x=1):

With those changes, your function actually works:

>>> Quadratic(1, 3, -4)
(1, -4)

Help: since the Square_Root function requires 2 parameters, what would I add inside `Sqrt_Disc = Square_Root(discriminant)` for Quadratic to work? — user1947570, Jan 04 '13 at 06:01
@user1947570: you mean you don't know how to use the recursive function `Square_Root()`? Tsk. tsk. — Martijn Pieters, Jan 04 '13 at 06:05

score 3 · Answer 3 · answered Jan 04 '13 at 05:54

3

You need to do 4 * (a * c) or just 4 * a * c because python thinks you are trying to call a function 4

answered Jan 04 '13 at 05:54

Volatility

31,232
10
80
89

TypeError: 'int' object is not callable: Quadratic

3 Answers3