Square Root Of Given Number In Java Without Built-in Functions

Chapter: Interview Programs Last Updated: 14-06-2016 14:03:07 UTC

Program:

            /* ............... START ............... */
                
public class JavaSquareRootWithOutBuiltInFunction {

	public static void main(String[] args) {
		double number = 36;
		findSquareRoot(number);

	}

	public static void findSquareRoot(double number) {

		boolean isPositiveNumber = true;
		double g1;

		if (number == 0) {
			System.out.println("Square root of " + number + " = " + 0);
		}

		else if (number < 0) {
			number = -number;
			isPositiveNumber = false;
		}

		double squareRoot = number / 2;
		do {
			g1 = squareRoot;
			squareRoot = (g1 + (number / g1)) / 2;
		} while ((g1 - squareRoot) != 0);

		if (isPositiveNumber) {
			System.out.println("Square roots of " + number + " are ");
			System.out.println("+" + squareRoot);
			System.out.println("-" + squareRoot);
		}

		else {
			System.out.println("Square roots of -" + number + " are ");
			System.out.println("+" + squareRoot + " i");
			System.out.println("-" + squareRoot + " i");
		}
	}
}
                /* ............... END ............... */
        

Output

Square roots of 36.0 are 
+6.0
-6.0

Tags

Square Root Of Given Number Without Built-in Functions, Java, Interview Programs

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