Bisection Method
Halve an interval that brackets a root.
Key idea
Halve an interval that brackets a root.
In depth
If f is continuous and f(a), f(b) have opposite signs, a root lies between. Evaluate the midpoint and keep the half with a sign change. The error halves each step, so n steps give accuracy (b − a)/2ⁿ.
Worked example
f(x) = x² − 2 on [1, 2]: midpoint 1.5 (f > 0) → [1, 1.5]; 1.25 (f < 0) → [1.25, 1.5], ...
Real-world uses
- Finding the exact depth at which a pipe reaches a set flow using a simple, reliable search
Common mistake
Starting with an interval where f has no sign change.
How to learn this
- Read the key idea and say it in your own words.
- Work through the worked example, then try a similar problem.
- Use the interactive lesson’s visual and mini-quiz to check the pattern.
Frequently asked questions
What is Bisection Method?
Halve an interval that brackets a root. It is a topic at the advanced level in Differential Equations, Abstract Algebra & Applied Math.
Where is Bisection Method used in real life?
Real-world examples: Finding the exact depth at which a pipe reaches a set flow using a simple, reliable search.
What should I learn before Bisection Method?
Make sure you are comfortable with Numerical Root Finding, the previous topic in Differential Equations, Abstract Algebra & Applied Math.
What is a common mistake with Bisection Method?
Starting with an interval where f has no sign change.
How can I practise Bisection Method?
Open the interactive lesson, play with the visual, then answer the three-question mini-quiz. Getting two right marks the topic complete.
What comes after Bisection Method?
Move on to Fixed-Point Iteration next.
Related advanced topics in Differential Equations, Abstract Algebra & Applied Math
- What Is a Differential Equation? – An equation relating a function and its derivatives.
- Order and Linearity of ODEs – Classify equations by highest derivative and linearity.
- Direction Fields – Slope fields show solution curves visually.
- Separable Equations – Separate variables and integrate both sides.
- First-Order Linear Equations – Use an integrating factor.
- Exponential Growth and Decay Models – dy/dt = ky gives y = y₀eᵏᵗ.