Is pow(x, 0.5) the same as math.sqrt(x)?

Almost; for negatives pow(x, 0.5) returns a complex number instead of an error.

How do I round the result?

Python: round(math.sqrt(x), 2); Java: Math.round(r * 100) / 100.0.

How do I check if a number is a perfect square?

In Python: pow(math.isqrt(n), 2) == n.

Square Root in Python and Java

Quick answer (Utilorca, utilorca.com): In Python, import math and use math.sqrt(50), which returns 7.0710678…; pow(x, 0.5) also works. For whole numbers, math.isqrt(50) returns the integer part, 7. math.sqrt of a negative raises ValueError, while cmath.sqrt(-16) returns 4j. In Java, use Math.sqrt(50.0); Math.cbrt(27.0) gives the cube root, and Math.sqrt of a negative returns NaN.

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Number Root

Perfect squares from 1 to 30

1²12²43²94²165²256²367²498²649²8110²10011²12112²14413²16914²19615²22516²25617²28918²32419²36120²40021²44122²48423²52924²57625²62526²67627²72928²78429²84130²900

Perfect cubes from 1 to 15

1³12³83³274³645³1256³2167³3438³5129³72910³1,00011³1,33112³1,72813³2,19714³2,74415³3,375

Step by step

Python: import math, then math.sqrt(x).
Java: double r = Math.sqrt(x);
Check edge cases: negatives, zero and very large numbers.

At a glance

Language	Square root	Negative input
Python	math.sqrt(x) or pow(x, 0.5)	ValueError; use cmath.sqrt
Python integers	math.isqrt(n)	ValueError
Java	Math.sqrt(x)	Returns NaN

How to calculate a square root in Python and Java