Home > Numerical methods > Roots of Non-Linear Equation > Secant method example

5. Secant method example ( Enter your problem )
  1. Algorithm & Example-1 `f(x)=x^3-x-1`
  2. Example-2 `f(x)=2x^3-2x-5`
  3. Example-3 `x=sqrt(12)`
  4. Example-4 `x=root(3)(48)`
  5. Example-5 `f(x)=x^3+2x^2+x-1`

2. Example-2 `f(x)=2x^3-2x-5`





Find a root of an equation `f(x)=2x^3-2x-5` using Secant method

Solution:
Here `2x^3-2x-5=0`

Let `f(x) = 2x^3-2x-5`

Here
`x`012
`f(x)`-5-57



`1^(st)` iteration :

`x_0 = 1` and `x_1 = 2`

`f(x_0) = f(1) = -5` and `f(x_1) = f(2) = 7`

`:. x_2 = x_0 - f(x_0) * (x_1 - x_0) / (f(x_1) - f(x_0))`

`x_2 = 1 - (-5) xx (2 - 1)/(7 - (-5))`

`x_2 = 1.41667`

`:. f(x_2)=f(1.41667)=2*1.41667^(3)-2*1.41667-5=-2.14699`


`2^(nd)` iteration :

`x_1 = 2` and `x_2 = 1.41667`

`f(x_1) = f(2) = 7` and `f(x_2) = f(1.41667) = -2.14699`

`:. x_3 = x_1 - f(x_1) * (x_2 - x_1) / (f(x_2) - f(x_1))`

`x_3 = 2 - 7 xx (1.41667 - 2)/(-2.14699 - 7)`

`x_3 = 1.55359`

`:. f(x_3)=f(1.55359)=2*1.55359^(3)-2*1.55359-5=-0.60759`


`3^(rd)` iteration :

`x_2 = 1.41667` and `x_3 = 1.55359`

`f(x_2) = f(1.41667) = -2.14699` and `f(x_3) = f(1.55359) = -0.60759`

`:. x_4 = x_2 - f(x_2) * (x_3 - x_2) / (f(x_3) - f(x_2))`

`x_4 = 1.41667 - (-2.14699) xx (1.55359 - 1.41667)/(-0.60759 - (-2.14699))`

`x_4 = 1.60763`

`:. f(x_4)=f(1.60763)=2*1.60763^(3)-2*1.60763-5=0.09449`


`4^(th)` iteration :

`x_3 = 1.55359` and `x_4 = 1.60763`

`f(x_3) = f(1.55359) = -0.60759` and `f(x_4) = f(1.60763) = 0.09449`

`:. x_5 = x_3 - f(x_3) * (x_4 - x_3) / (f(x_4) - f(x_3))`

`x_5 = 1.55359 - (-0.60759) xx (1.60763 - 1.55359)/(0.09449 - (-0.60759))`

`x_5 = 1.60036`

`:. f(x_5)=f(1.60036)=2*1.60036^(3)-2*1.60036-5=-0.00324`


`5^(th)` iteration :

`x_4 = 1.60763` and `x_5 = 1.60036`

`f(x_4) = f(1.60763) = 0.09449` and `f(x_5) = f(1.60036) = -0.00324`

`:. x_6 = x_4 - f(x_4) * (x_5 - x_4) / (f(x_5) - f(x_4))`

`x_6 = 1.60763 - 0.09449 xx (1.60036 - 1.60763)/(-0.00324 - 0.09449)`

`x_6 = 1.6006`

`:. f(x_6)=f(1.6006)=2*1.6006^(3)-2*1.6006-5=-0.00002`


Approximate root of the equation `2x^3-2x-5=0` using Secant method is `1.6006`

`n``x_0``f(x_0)``x_1``f(x_1)``x_2``f(x_2)`Update
11-5271.41667-2.14699`x_0 = x_1`
`x_1 = x_2`
2271.41667-2.146991.55359-0.60759`x_0 = x_1`
`x_1 = x_2`
31.41667-2.146991.55359-0.607591.607630.09449`x_0 = x_1`
`x_1 = x_2`
41.55359-0.607591.607630.094491.60036-0.00324`x_0 = x_1`
`x_1 = x_2`
51.607630.094491.60036-0.003241.6006-0.00002`x_0 = x_1`
`x_1 = x_2`





This material is intended as a summary. Use your textbook for detail explanation.
Any bug, improvement, feedback then Submit Here





Share this solution or page with your friends.
 
 
Copyright © 2026. All rights reserved. Terms, Privacy
 
 

.