Option 1: math.sqrt()

The math module from the standard library has a sqrt function to calculate the square root of a number. It takes any type that can be converted to float (which includes int) and returns a float.

Use with AI>>> import math
>>> math.sqrt(9)
3.0

Option 2: Fractional exponent

The power operator (**) or the built-in pow() function can also be used to calculate a square root. Mathematically speaking, the square root of a equals a to the power of 1/2.

The power operator requires numeric types and matches the conversion rules for binary arithmetic operators, so in this case it will return either a float or a complex number.

Use with AI>>> 9 ** (1/2)
3.0
>>> 9 ** .5  # Same thing
3.0
>>> 2 ** .5
1.4142135623730951

(Note: in Python 2, 1/2 is truncated to 0, so you have to force floating point arithmetic with 1.0/2 or similar. See Why does Python give the "wrong" answer for square root?)

This method can be generalized to nth root, though fractions that can't be exactly represented as a float (like 1/3 or any denominator that's not a power of 2) may cause some inaccuracy:

Use with AI>>> 8 ** (1/3)
2.0
>>> 125 ** (1/3)
4.999999999999999

Edge cases

Negative and complex

Exponentiation works with negative numbers and complex numbers, though the results have some slight inaccuracy:

Use with AI>>> (-25) ** .5  # Should be 5j
(3.061616997868383e-16+5j)
>>> 8j ** .5  # Should be 2+2j
(2.0000000000000004+2j)

(Note: the parentheses are required on -25, otherwise it's parsed as -(25**.5) because exponentiation is more tightly binding than negation.)

Meanwhile, math is only built for floats (which are real numbers), so for x<0, math.sqrt(x) will raise ValueError: math domain error and for complex x, it'll raise TypeError: can't convert complex to float. Instead, you can use cmath.sqrt(x), which is more accurate than exponentiation (and will likely be faster too):

Use with AI>>> import cmath
>>> cmath.sqrt(-25)
5j
>>> cmath.sqrt(8j)
(2+2j)

Precision

Both options involve an implicit conversion to float, so floating point precision is a factor. For example let's try a big number:

Use with AI>>> n = 10**30
>>> x = n**2
>>> root = x**.5
>>> root == n
False
>>> root - n  # how far off are they?
0.0
>>> int(root) - n  # how far off is the float from the int?
19884624838656

Very large numbers might not even fit in a float and you'll get OverflowError: int too large to convert to float. See Python sqrt limit for very large numbers?

Other types

Let's look at Decimal for example:

Exponentiation fails unless the exponent is also Decimal:

Use with AI>>> decimal.Decimal('9') ** .5
Traceback (most recent call last):
  File "<stdin>", line 1, in <module>
TypeError: unsupported operand type(s) for ** or pow(): 'decimal.Decimal' and 'float'
>>> decimal.Decimal('9') ** decimal.Decimal('.5')
Decimal('3.000000000000000000000000000')

Meanwhile, math and cmath will silently convert their arguments to float and complex respectively, which could mean loss of precision.

decimal also has its own .sqrt(). See also calculating n-th roots using Python 3's decimal module

You could use the Babylonian method.

def sqrt(x):
    n = 1
    for _ in range(10):
        print(n)
        n = (n + x/n) * 0.5

It converges extremely fast. Here's an example for sqrt(2):

1
1.5
1.41666666667
1.41421568627
1.41421356237
1.41421356237
1.41421356237
1.41421356237
1.41421356237
1.41421356237

for sqrt(3):

1
2.0
1.75
1.73214285714
1.73205081001
1.73205080757
1.73205080757
1.73205080757
1.73205080757
1.73205080757

Now you just need to replace the for loop with a while and a precision condition and return the result instead of just printing it.