---
title: "Bisection method working principle"
subject: "Numerical Method"
chapter: 2
key: "semester_4/numerical_method:bisection-working-principle"
source: https://www.bctengineer.com/q/semester_4/numerical_method/bisection-working-principle
site: BCT Engineering
---

# Bisection method working principle

Numerical Method · Chapter 2: Solutions of Nonlinear Equations

- Asked in: 2068 Bhadra, 4 marks
- Video: [Bisection method | solution of non linear algebraic equation](https://www.youtube.com/watch?v=qEecNyRa5o4) — Smart Engineer
- Answer checked by Claude Opus 5.5 (high effort) on 2026-10-14
- Page: https://www.bctengineer.com/q/semester_4/numerical_method/bisection-working-principle
- Whole chapter: https://www.bctengineer.com/md/semester_4/numerical_method/chapter-2

## Question

Explain the working principle of the bisection method to find a real root of a non-linear equation.

## Short answer

Bisection uses the intermediate value theorem: if f(a) and f(b) have opposite signs, the root lies between them. At each step take c=(a+b)/2, keep the half interval where the sign change remains, and repeat until the interval is smaller than tolerance. It always converges linearly and needs only function values.

## Answer

The bisection method is a bracketing method based on the **Intermediate Value Theorem**: if f(x) is continuous on [a, b] and f(a)·f(b) < 0, then f(x) = 0 has at least one real root between a and b.

#### Working principle
1. Choose a and b such that f(a) and f(b) have opposite signs.
2. Find the midpoint c = (a + b)/2.
3. Test the sign of f(c):
   - if f(c) = 0, c is the root;
   - if f(a)·f(c) < 0, the root lies in [a, c], so set b = c;
   - otherwise the root lies in [c, b], so set a = c.
4. The new interval is exactly half as long as the old one and still contains the root. Repeat from step 2.
5. Stop when the interval length (b − a) is less than the tolerance, or |f(c)| is very small. The last midpoint is the root.

```text
 a                  c                  b
 |------------------|------------------|
 f(a)<0           f(c)>0             f(b)>0
 root is in [a, c]  ->  set b = c
```

#### Error and iterations
After n steps the interval length is (b − a)/2ⁿ, so the error is at most (b − a)/2ⁿ. For tolerance e the number of iterations needed is

n ≥ [log(b − a) − log(e)] / log 2

Example: for f(x) = x² − 2 on [1, 2], c = 1.5 gives f(c) = 0.25 > 0 and f(1) < 0, so the root is in [1, 1.5]; next c = 1.25, and so on toward √2 = 1.4142.

#### Features
- Always converges (if the initial interval is valid), but slowly (linear convergence).
- Needs only function values, not derivatives.
- Cannot find roots where the curve touches the axis without crossing (even multiplicity).

## Check yourself

The answer's key points as multiple-choice questions, with the correct option and why.

1. Bisection relies on:
   A. Intermediate Value Theorem: continuity and f(a)f(b) < 0
   B. Taylor series
   C. Mean value of derivative
   D. Rolle for f'
   Answer: A. Sign change implies a root. (Key point: Theorem)

2. After each step the interval:
   A. Doubles
   B. Halves while still containing the root
   C. Stays the same
   D. Shifts only
   Answer: B. New length is half. (Key point: Interval)

3. After n steps error is at most:
   A. (b-a)*n
   B. (b-a)/n
   C. (b-a)/2^n
   D. (b-a)^n
   Answer: C. Halved each step. (Key point: Error)

4. Bisection cannot find roots where:
   A. The curve crosses
   B. f is continuous
   C. Sign changes
   D. The curve touches the axis without crossing
   Answer: D. There is no sign change at even multiplicity. (Key point: Limitation)

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