Secant method to solve non-linear equation

Data StructureMathematical ProblemsAlgorithms

Secant method is also used to solve non-linear equations. This method is similar to the Newton-Raphson method, but here we do not need to find the differentiation of the function f(x). Only using f(x), we can find f’(x) numerically by using Newton’s Divide difference formula. From the Newton-Raphson formula,

we know that,

Now, using divide difference formula, we get,

By replacing the f’(x) of Newton-Raphson formula by the new f’(x), we can find the secant formula to solve non-linear equations.

Note: For this method, we need any two initial guess to start finding the root of non-linear equations.

Input and Output

The function f(x) = (x*x) - (4*x) - 10
The root is: -1.74166


secant(x1, x2)

Input: Two initial guess for root.

Output: The approximate root of a non-linear equation f(x).

   f1 := f(x1)
   f2 := f(x2)
   x3 := ((f2*x1) – (f1*x2)) / (f2 – f1)
   while relative error of x3 and x2 are > precision, do
      x1 := x2
      f1 := f2
      x2 := x3
      f2 := f(x2)
      x3 := ((f2*x1) – (f1*x2)) / (f2 – f1)
   root := x3
   return root


using namespace std;

double absolute(double value) {             //to find magnitude of value
   if(value < 0)
      return (-value);
   return value;

double f(double x) {              //the given function x^2-4x-10
   return ((x*x)-(4*x)-10);

double secant(double x1, double x2) {
   double x3, root;
   double f1, f2;
   f1 = f(x1);
   f2 = f(x2);
   x3 = (f2*x1-f1*x2)/(f2-f1);

   while(absolute((x3-x2)/x3) > 0.00001) {         //test accuracy of x3
      x1 = x2;           //shift x values
      f1 = f2;
      x2 = x3;
      f2 = f(x2);                 //find new x2
      x3 = (f2*x1-f1*x2)/(f2-f1);          //calculate x3

   root = x3;
   return root;              //root of the equation

main() {
   double a, b, res;
   a = 0.5;
   b = 0.75;
   res = secant(a, b);
   cout << "The root is: " << res;


The root is: -1.74166
Published on 11-Jul-2018 13:27:35