// C++ Program to find root of an  // equations using secant method #include    using namespace std; // function takes value of x and returns f(x) float f(float x) {  // we are taking equation as x^3+x-1  float f = pow(x 3) + x - 1;  return f; } void secant(float x1 float x2 float E) {  float n = 0 xm x0 c;  if (f(x1) * f(x2) < 0) {  do {  // calculate the intermediate value  x0 = (x1 * f(x2) - x2 * f(x1)) / (f(x2) - f(x1));  // check if x0 is root of equation or not  c = f(x1) * f(x0);  // update the value of interval  x1 = x2;  x2 = x0;  // update number of iteration  n++;  // if x0 is the root of equation then break the loop  if (c == 0)  break;  xm = (x1 * f(x2) - x2 * f(x1)) / (f(x2) - f(x1));  } while (fabs(xm - x0) >= E); // repeat the loop  // until the convergence  cout << 'Root of the given equation=' << x0 << endl;  cout << 'No. of iterations = ' << n << endl;  } else  cout << 'Can not find a root in the given interval'; } // Driver code int main() {  // initializing the values  float x1 = 0 x2 = 1 E = 0.0001;  secant(x1 x2 E);  return 0; } 

// Java Program to find root of an  // equations using secant method class GFG {    // function takes value of x and   // returns f(x)  static float f(float x) {    // we are taking equation   // as x^3+x-1  float f = (float)Math.pow(x 3)   + x - 1;    return f;  }    static void secant(float x1 float x2  float E) {    float n = 0 xm x0 c;  if (f(x1) * f(x2) < 0)   {  do {    // calculate the intermediate  // value  x0 = (x1 * f(x2) - x2 * f(x1))  / (f(x2) - f(x1));    // check if x0 is root of  // equation or not  c = f(x1) * f(x0);    // update the value of interval  x1 = x2;  x2 = x0;    // update number of iteration  n++;    // if x0 is the root of equation   // then break the loop  if (c == 0)  break;  xm = (x1 * f(x2) - x2 * f(x1))   / (f(x2) - f(x1));    // repeat the loop until the   // convergence   } while (Math.abs(xm - x0) >= E);     System.out.println('Root of the' +  ' given equation=' + x0);    System.out.println('No. of '  + 'iterations = ' + n);  }     else  System.out.print('Can not find a'  + ' root in the given interval');  }    // Driver code  public static void main(String[] args) {    // initializing the values  float x1 = 0 x2 = 1 E = 0.0001f;  secant(x1 x2 E);  } } // This code is contributed by Anant Agarwal. 

# Python3 Program to find root of an  # equations using secant method  # function takes value of x  # and returns f(x)  def f(x): # we are taking equation  # as x^3+x-1  f = pow(x 3) + x - 1; return f; def secant(x1 x2 E): n = 0; xm = 0; x0 = 0; c = 0; if (f(x1) * f(x2) < 0): while True: # calculate the intermediate value  x0 = ((x1 * f(x2) - x2 * f(x1)) / (f(x2) - f(x1))); # check if x0 is root of  # equation or not  c = f(x1) * f(x0); # update the value of interval  x1 = x2; x2 = x0; # update number of iteration  n += 1; # if x0 is the root of equation  # then break the loop  if (c == 0): break; xm = ((x1 * f(x2) - x2 * f(x1)) / (f(x2) - f(x1))); if(abs(xm - x0) < E): break; print('Root of the given equation =' round(x0 6)); print('No. of iterations = ' n); else: print('Can not find a root in ' 'the given interval'); # Driver code  # initializing the values  x1 = 0; x2 = 1; E = 0.0001; secant(x1 x2 E); # This code is contributed by mits 

// C# Program to find root of an  // equations using secant method using System; class GFG {    // function takes value of   // x and returns f(x)  static float f(float x)   {    // we are taking equation   // as x^3+x-1  float f = (float)Math.Pow(x 3)   + x - 1;  return f;  }    static void secant(float x1 float x2  float E)     {    float n = 0 xm x0 c;  if (f(x1) * f(x2) < 0)   {  do {    // calculate the intermediate  // value  x0 = (x1 * f(x2) - x2 * f(x1))  / (f(x2) - f(x1));    // check if x0 is root of  // equation or not  c = f(x1) * f(x0);    // update the value of interval  x1 = x2;  x2 = x0;    // update number of iteration  n++;    // if x0 is the root of equation   // then break the loop  if (c == 0)  break;  xm = (x1 * f(x2) - x2 * f(x1))   / (f(x2) - f(x1));    // repeat the loop until   // the convergence   } while (Math.Abs(xm - x0) >= E);     Console.WriteLine('Root of the' +  ' given equation=' + x0);    Console.WriteLine('No. of ' +   'iterations = ' + n);  }     else  Console.WriteLine('Can not find a' +   ' root in the given interval');  }    // Driver code  public static void Main(String []args)   {    // initializing the values  float x1 = 0 x2 = 1 E = 0.0001f;  secant(x1 x2 E);  } } // This code is contributed by vt_m. 

 // PHP Program to find root of an  // equations using secant method // function takes value of x  // and returns f(x) function f( $x) { // we are taking equation // as x^3+x-1 $f = pow($x 3) + $x - 1; return $f; } function secant($x1 $x2 $E) { $n = 0; $xm; $x0; $c; if (f($x1) * f($x2) < 0) { do { // calculate the intermediate value $x0 = ($x1 * f($x2) - $x2 * f($x1)) / (f($x2) - f($x1)); // check if x0 is root  // of equation or not $c = f($x1) * f($x0); // update the value of interval $x1 = $x2; $x2 = $x0; // update number of iteration $n++; // if x0 is the root of equation // then break the loop if ($c == 0) break; $xm = ($x1 * f($x2) - $x2 * f($x1)) / (f($x2) - f($x1)); // repeat the loop // until the convergence } while (abs($xm - $x0) >= $E); echo 'Root of the given equation='. $x0.'n' ; echo 'No. of iterations = '. $n ; } else echo 'Can not find a root in the given interval'; } // Driver code { // initializing the values $x1 = 0; $x2 = 1; $E = 0.0001; secant($x1 $x2 $E); return 0; } // This code is contributed by nitin mittal. ?> 

<script> // JavaScript Program to find root of an // equations using secant method // function takes value of x and returns f(x) function f(x) {  // we are taking equation as x^3+x-1  let f = Math.pow(x 3) + x - 1;  return f; } function secant(x1 x2 E) {  let n = 0 xm x0 c;  if (f(x1) * f(x2) < 0) {  do {  // calculate the intermediate value  x0 = (x1 * f(x2) - x2 * f(x1)) / (f(x2) - f(x1));  // check if x0 is root of equation or not  c = f(x1) * f(x0);  // update the value of interval  x1 = x2;  x2 = x0;  // update number of iteration  n++;  // if x0 is the root of equation then break the loop  if (c == 0)  break;  xm = (x1 * f(x2) - x2 * f(x1)) / (f(x2) - f(x1));  } while (Math.abs(xm - x0) >= E); // repeat the loop  // until the convergence  document.write('Root of the given equation=' + x0.toFixed(6) + '  
');  document.write('No. of iterations = ' + n + '  
');  } else  document.write('Can not find a root in the given interval'); } // Driver code  // initializing the values  let x1 = 0 x2 = 1 E = 0.0001;  secant(x1 x2 E); // This code is contributed by Surbhi Tyagi. </script>