#include <stdio.h>
#include <stdlib.h>
#include <string.h>
#include <math.h>

int main(int argc, char **argv)
{
   long double k;
   long double k2;
   long double k3;
   long double k4;
   long double k5;
   long double ans;
   long double x;
   long double x10;
   long double x8;
   long double x6;
   long double x4;
   long double x2;
   char pszString[256] = "2";

   if ( argc < 2 )
   {
      printf( "Enter a number to find square root of:" );
      fgets( pszString, sizeof( pszString ), stdin );
      puts( pszString );
  }
   else
      strncpy( pszString, argv[1], sizeof( pszString ) );

   k = atof( pszString );
   k2 = k*k;
   k3 = k2*k;
   k4 = k3*k;
   k5 = k4*k;
   x = sqrtl( k ) * 1.1; /* form estimate correct to 1 digit */
   x2 = x*x;
   x4 = x2*x2;
   x6 = x4*x2;
   x8 = x6*x2;
   x10= x8*x2;
   /* Method A, term = 9 */
   ans = ( x10 + 45*x8*k + 210*x6*k2 + 210*x4*k3 + 45*x2*k4 + k5) /
         ( 2*x*( 5*x8 + 60*x6*k + 126*x4*k2 + 60*x2*k3 + 5*k4 ) );
   printf("answer = %30.20LeL\nx      = %30.20LeL\n", sqrtl( k ), ans );

   /* Method B, term = 9 */
   ans = ( 2*x*k*( 5*x8 + 60*x6*k + 126*x4*k2 + 60*x2*k3 + 5*k4 ) ) /
         ( x10 + 45*x8*k + 210*x6*k2 + 210*x4*k3 + 45*x2*k4 + k5 ) ;
   printf("answer = %30.20LeL\nx      = %30.20LeL\n", sqrtl( k ), ans );
   return 0;
}