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One of the most popular frequently asked Java Interview Question is, “Given an integer x, write a java program to find the square root of it”. There are many ways to solve this problem. In this article, let’s check out different ways to find square and square root in Java.
Before discussing the square root code in Java, let’s understand the term square root first.
The square of a number is that number times itself. In other terms, when we multiply an integer by itself we call the product the square of the number. Mathematically, the square of a number is given as,
Square of n = n*n
For example, the square of number 4 is 4*4 = 16
The square root is just the opposite of the square. The square root of a number, n, is the number that gives n when multiplied by itself. Mathematically, the square root of a number is given as,
Now that you know what square and square root of a number are, let’s see different ways to calculate them in Java.
You can square a number in Java in two different ways:
Here’s a Java Program to square a number by multiplying it by itself.
package MyPackage; import java.util.Scanner; public class Square1 { public static void main(String args[]) { Double num; Scanner sc= new Scanner(System.in); System.out.print("Enter a number: "); num=sc.nextDouble(); Double square = num*num; System.out.println("Square of "+ num + " is: "+ square); } }
Enter a number: 10 Square of 10.0 is: 100.0
Here’s a Java Program to call the Math.pow method to square a number.
package MyPackage; import java.util.Scanner; import java.lang.Math; public class Square2 { public static void main(String args[]) { Double num; Scanner sc= new Scanner(System.in); System.out.print("Enter a number: "); num = sc.nextDouble(); Double square = Math.pow(num, 2); System.out.println("Square of "+ num + " is: "+ square); } }
Output
Enter a number: 22 Square of 22.0 is: 484.0
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Now let’s check out how to calculate the square root of a number in Java.
There are multiple ways to find square root a given number in Java. Let’s explore a few of those.
Syntax
public static double sqrt(double x)
Code
package MyPackage; public class SquareRoot2 { public static void main(String args[]) { double a = 100; System.out.println(Math.sqrt(a)); // Input positive value, Output square root of x double b = -81.00; System.out.println(Math.sqrt(b)); // Input negative value, Output NaN double c = 0.0/0; // Input NaN, Output NaN System.out.println(Math.sqrt(c)); double d = 1.0/0; // Input positive infinity, Output positive infinity System.out.println(Math.sqrt(d)); double e = 0.0; // Input positive Zero, Output positive zero System.out.println(Math.sqrt(e)); } }
10.0 NaN NaN Infinity 0.0
We can use the logic √number = number½ to find the square root of a number.
Code
package MyPackage; import java.util.Scanner; public class SquareRoot1 { public static void main(String[] args) { Double num; Scanner sc= new Scanner(System.in); System.out.print("Enter a number: "); num = sc.nextDouble(); Double squareroot = Math.pow(num, 0.5); System.out.println("The Square of a Given Number " + num + " = " + squareroot); } }
Enter a number: 81 The Square of a Given Number 81.0 = 9.0
Here’s the logic that we are using:
The first sqrt number should be the input number / 2. Here’s a Java Program implementing the above logic.
Code
package MyPackage; public class SquareRoot { public static double square(double number){ double t; double squareroot = number / 2; do { t = squareroot; squareroot = (t + (number / t)) / 2; } while ((t - squareroot) != 0); return squareroot; } public static void main(String[] args) { double number = 16; double root; root = square(number); System.out.println("Number : "+number); System.out.println("Square Root : "+root); } }
Output
Number : 121.0 Square Root : 11.0
This brings us to the end of this article.
Make sure you practice as much as possible and revert your experience.
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