- Interval Bisection

With this method we repeatedly bisect the interval containing the root in order to improve the approximation.

Given that \(x^3-3x^2=1-x\) has a root between 2 and 3, find this root to 1 decimal place using the bisection method.
Rewrite the equation in the form \(f(x)=0\)
Let \(f(x)=x^3-3x^2+x-1\)
  • Determine the midpoint of this interval in which the root lies.
Mid – point of interval is 2.5
  • Since there is a sign change between 2.5 and 3 the root occurs within this interval. We therefore now bisect this interval. Mid – point is 2.75
\[ =-0.140625\]
  • Due to sign changes the root must be in the interval 2.75 and 3.
Mid – point is 2.875
  • Due to sign changes the root is between 2.75 and 2.875
Mid – point is 2.8125
Therefore, root lies between 2.75 and 2.8125. When these 2 values are rounded to 1 decimal place the answers are both 2.8. Consequently, the root is approximately 2.8.