Java Program to Find Square Root of a Number

You can find the square root of a number in Java using Math.sqrt(number), for example: double root = Math.sqrt(25); which returns 5.0.

Examples

Input25

Output5.0

Input0

Output0.0

Input2

Output1.4142135623730951

How to Think About It

To find the square root, think about what number multiplied by itself gives the original number. We use Java's built-in Math.sqrt() method which calculates this value directly and accurately.

Algorithm

1

Get the input number from the user.

2

Use the built-in function to calculate the square root.

3

Display the result to the user.

Code

java

import java.util.Scanner;

public class SquareRoot {
    public static void main(String[] args) {
        Scanner scanner = new Scanner(System.in);
        System.out.print("Enter a number: ");
        double number = scanner.nextDouble();
        double root = Math.sqrt(number);
        System.out.println("Square root of " + number + " is " + root);
        scanner.close();
    }
}

Output

Enter a number: 25
Square root of 25.0 is 5.0

Dry Run

Let's trace the input 25 through the code

1

Input number

User enters 25, stored in variable number = 25.0

2

Calculate square root

Call Math.sqrt(25.0), result is 5.0 stored in root

3

Print result

Prints: Square root of 25.0 is 5.0

Step	Variable	Value
1	number	25.0
2	root	5.0
3	output	Square root of 25.0 is 5.0

Why This Works

Step 1: Input reading

We read the number from the user using Scanner to get the value to find the square root of.

Step 2: Square root calculation

We use Math.sqrt() which is a built-in Java method that returns the square root of the given number.

Step 3: Output display

We print the result so the user can see the square root value.

Alternative Approaches

Using exponentiation operator

java

import java.util.Scanner;

public class SquareRoot {
    public static void main(String[] args) {
        Scanner scanner = new Scanner(System.in);
        System.out.print("Enter a number: ");
        double number = scanner.nextDouble();
        double root = Math.pow(number, 0.5);
        System.out.println("Square root of " + number + " is " + root);
        scanner.close();
    }
}

Uses <code>Math.pow()</code> with 0.5 exponent to find square root; slightly less direct but works the same.

Using Newton's method (manual calculation)

java

import java.util.Scanner;

public class SquareRoot {
    public static double sqrtNewton(double n) {
        double x = n;
        double y = 1;
        double e = 0.000001; // error tolerance
        while (x - y > e) {
            x = (x + y) / 2;
            y = n / x;
        }
        return x;
    }

    public static void main(String[] args) {
        Scanner scanner = new Scanner(System.in);
        System.out.print("Enter a number: ");
        double number = scanner.nextDouble();
        double root = sqrtNewton(number);
        System.out.println("Square root of " + number + " is " + root);
        scanner.close();
    }
}

Manually calculates square root using Newton's iterative method; useful for learning but more complex.

Complexity: O(1) time, O(1) space

Time Complexity

The calculation uses a built-in method that runs in constant time, so it is O(1).

Space Complexity

Only a few variables are used, so space complexity is O(1).

Which Approach is Fastest?

Using Math.sqrt() is fastest and simplest compared to manual methods like Newton's method.

Approach	Time	Space	Best For
Math.sqrt()	O(1)	O(1)	Quick and accurate calculation
Math.pow(number, 0.5)	O(1)	O(1)	Alternative using exponentiation
Newton's method	O(k) where k is iterations	O(1)	Learning manual calculation, customizable precision