text: Babylonian method for square root |  ; Returns the square root of n . Note that the function ; We are using n itself as initial approximation This can definitely be improved ; e decides the accuracy level ; Driver program to test above function
code: class GFG { static float squareRoot ( float n ) { float x = n ; float y = 1 ; double e = 0.000001 ; while ( x - y > e ) { x = ( x + y ) / 2 ; y = n / x ; } return x ; } public static void main ( String [ ] args ) { int n = 50 ; System . out . printf ( " Square ▁ root ▁ of ▁ " + n + " ▁ is ▁ " + squareRoot ( n ) ) ; } }