The question is:

The following method was known to the ancient Greeks for computing square roots. Given a value x > 0 and a guess g for the square root, a better guess is (x + g/x) / 2 (g + x/g) / 2. Write a recursive helper method public static squareRootGuess(double x, double g). If g2 is approximately equal to x, return g, otherwise, return squareRootGuess with the better guess. Then write a method public static squareRoot(double x) that uses the helper method.

So, so far I've got:

Code :

public class main1c { public static void main(String[] args){ Scanner keys = new Scanner(System.in); } public static double test(double x, double g) { if (closeEnough(x/g, g)) return g; else return test(g+x/g) } static boolean closeEnough(double a, double b) { return (Math.abs(a - b) < (b * 0.1)); // a is within 1% of b } }

I'm not sure where to go from here. How do I make a recursion method with the equation(g + x/g) / 2?