Cubic root of 3 is 1.44225 26Cubic root of 3 is 1.44225 27Cubic root of 3 is 1.44225 28Cubic root of 3 is 1.44225 29Cubic root of 3 is 1.44225 30
OutputCubic root of 3 is 1.44225
Time Complexity: O(logn)
Auxiliary Space: O(1) Please refer complete article on Find cubic root of a number for more details!
Method #2: Using power(**) function
Python3
Cubic root of 3 is 1.44225 31
Cubic root of 3 is 1.44225 16
# of a number using Binary Search2# Python 3 program to find cubic root7 Cubic root of 3 is 1.44225 19
Cubic root of 3 is 1.44225 20# Returns the absolute value of1Cubic root of 3 is 1.44225 22Cubic root of 3 is 1.44225 23Cubic root of 3 is 1.44225 24Cubic root of 3 is 1.44225 25Cubic root of 3 is 1.44225 27Cubic root of 3 is 1.44225 77Cubic root of 3 is 1.44225 29Cubic root of 3 is 1.44225 30
In this program, we store the number in num and find the square root using the ** exponent operator. This program works for all positive real numbers. But for negative or complex numbers, it can be done as follows.
Source code: For real or complex numbers
# Find square root of real or complex numbers # Importing the complex math module import cmath num = 1+2j # To take input from the user #num = eval(input('Enter a number: ')) num_sqrt = cmath.sqrt(num) print('The square root of {0} is {1:0.3f}+{2:0.3f}j'.format(num ,num_sqrt.real,num_sqrt.imag))Output
The square root of (1+2j) is 1.272+0.786jIn this program, we use the sqrt() function in the cmath (complex math) module.
Note: If we want to take complex number as input directly, like 3+4j, we have to use the eval() function instead of float().
The eval() method can be used to convert complex numbers as input to the The square root of 8.000 is 2.828 1 objects in Python. To learn more, visit Python eval() function.
A Python list contains two positive integers. Write a Python program to check whether the cube root of the first number is equal to the square root of the second number.
Sample Data:
([8, 4]) -> True
([64, 16]) -> True
([64, 36]) -> False
Sample Solution-1:
Python Code:
def test(nums): x = nums[0] y = nums[1] t = y**0.5 if(x == t*t*t): return True else: return False nums = [8, 4] print("Original list of positive numbers:") print(nums) print(test(nums)) print("Check square root and cube root of the said numbers:") nums = [64, 16] print("Original list of positive numbers:") print(nums) print("Check square root and cube root of the said numbers:") print(test(nums)) nums = [64, 36] print("Original list of positive numbers:") print(nums) print("Check square root and cube root of the said numbers:") print(test(nums))Sample Output:
Original list of positive numbers: [8, 4] True Check square root and cube root of the said numbers: Original list of positive numbers: [64, 16] Check square root and cube root of the said numbers: True Original list of positive numbers: [64, 36] Check square root and cube root of the said numbers: FalseFlowchart:
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Are you trying to solve a quadratic equation? Maybe you need to calculate the length of one side of a right triangle. For these types of equations and more, the Python square root function, sqrt(), can help you quickly and accurately calculate your solutions.
By the end of this article, you’ll learn:
- What a square root is
- How to use the Python square root function, sqrt()
- When sqrt() can be useful in the real world
Let’s dive in!
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Square Roots in Mathematics
In algebra, a square, x, is the result of a number, n, multiplied by itself: x = n²
You can calculate squares using Python:
>>>>>> n = 5 >>> x = n ** 2 >>> x 25
The Python >>> 1 ** 2 1 >>> 2 ** 2 4 >>> 3 ** 2 9 2 operator is used for calculating the power of a number. In this case, 5 squared, or 5 to the power of 2, is 25.
The square root, then, is the number n, which when multiplied by itself yields the square, x.
In this example, n, the square root, is 5.
25 is an example of a perfect square. Perfect squares are the squares of integer values:
>>>>>> 1 ** 2 1 >>> 2 ** 2 4 >>> 3 ** 2 9
You might have memorized some of these perfect squares when you learned your multiplication tables in an elementary algebra class.
If you’re given a small perfect square, it may be straightforward enough to calculate or memorize its square root. But for most other squares, this calculation can get a bit more tedious. Often, an estimation is good enough when you don’t have a calculator.
Fortunately, as a Python developer, you do have a calculator, namely the !
Remove adsThe Python Square Root Function
Python’s >>> 1 ** 2 1 >>> 2 ** 2 4 >>> 3 ** 2 9 3 module, in the standard library, can help you work on math-related problems in code. It contains many useful functions, such as >>> 1 ** 2 1 >>> 2 ** 2 4 >>> 3 ** 2 9 4 and >>> 1 ** 2 1 >>> 2 ** 2 4 >>> 3 ** 2 9 5. It also includes the .
You’ll begin by importing >>> 1 ** 2 1 >>> 2 ** 2 4 >>> 3 ** 2 9 3:
>>>>>> import math
That’s all it takes! You can now use >>> 1 ** 2 1 >>> 2 ** 2 4 >>> 3 ** 2 9 8 to calculate square roots.
sqrt() has a straightforward interface.
It takes one parameter, >>> import math 0, which (as you saw before) stands for the square for which you are trying to calculate the square root. In the example from earlier, this would be >>> import math 1.
The return value of sqrt() is the square root of >>> import math 0, as a . In the example, this would be >>> import math 4.
Let’s take a look at some examples of how to (and how not to) use sqrt().
The Square Root of a Positive Number
One type of argument you can pass to sqrt() is a positive number. This includes both and types.
For example, you can solve for the square root of >>> import math 9 using sqrt():
>>>>>> math.sqrt(49) 7.0
The return value is >>> math.sqrt(49) 7.0 1 (the square root of >>> import math 9) as a floating point number.
Along with integers, you can also pass >>> import math 8 values:
>>>>>> math.sqrt(70.5) 8.396427811873332
You can verify the accuracy of this square root by calculating its inverse:
>>>>>> 8.396427811873332 ** 2 70.5
The Square Root of Zero
Even >>> math.sqrt(49) 7.0 4 is a valid square to pass to the Python square root function:
>>>>>> math.sqrt(0) 0.0
While you probably won’t need to calculate the square root of zero often, you may be passing a variable to sqrt() whose value you don’t actually know. So, it’s good to know that it can handle zero in those cases.
Remove adsThe Square Root of Negative Numbers
The square of any real number cannot be negative. This is because a negative product is only possible if one factor is positive and the other is negative. A square, by definition, is the product of a number and itself, so it’s impossible to have a negative real square:
>>>>>> math.sqrt(-25) Traceback (most recent call last): File "<stdin>", line 1, in <module> ValueError: math domain error
If you attempt to pass a negative number to sqrt(), then you’ll get a >>> math.sqrt(49) 7.0 7 because negative numbers are not in the domain of possible real squares. Instead, the square root of a negative number would need to be complex, which is outside the scope of the Python square root function.
Square Roots in the Real World
To see a real-world application of the Python square root function, let’s turn to the sport of tennis.
Imagine that Rafael Nadal, one of the fastest players in the world, has just hit a forehand from the back corner, where the baseline meets the sideline of the tennis court:
Now, assume his opponent has countered with a drop shot (one that would place the ball short with little forward momentum) to the opposite corner, where the other sideline meets the net:
How far must Nadal run to reach the ball?
You can determine from that the baseline is 27 feet long, and the sideline (on one side of the net) is 39 feet long. So, essentially, this boils down to solving for the hypotenuse of a right triangle:
Using a valuable equation from geometry, the Pythagorean theorem, we know that a² + b² = c², where a and b are the legs of the right triangle and c is the hypotenuse.
Therefore, we can calculate the distance Nadal must run by rearranging the equation to solve for c:
You can solve this equation using the Python square root function:
>>>>>> a = 27 >>> b = 39 >>> math.sqrt(a ** 2 + b ** 2) 47.43416490252569
So, Nadal must run about 47.4 feet (14.5 meters) in order to reach the ball and save the point.
Conclusion
Congratulations! You now know all about the Python square root function.
You’ve covered:
- A brief introduction to square roots
- The ins and outs of the Python square root function, sqrt()
- A practical application of sqrt() using a real-world example
Knowing how to use sqrt() is only half the battle. Understanding when to use it is the other. Now, you know both, so go and apply your newfound mastery of the Python square root function!
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Watch Now This tutorial has a related video course created by the Real Python team. Watch it together with the written tutorial to deepen your understanding: The Square Root Function in Python
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About Alex Ronquillo
Alex Ronquillo is a Software Engineer at thelab. He’s an avid Pythonista who is also passionate about writing and game development.
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