Exams
Test Series
Previous Year Papers
JEE Main Previous Year Question Paper JEE Advanced Previous Year Papers NEET Previous Year Question Paper CUET Previous Year Papers COMEDK UGET Previous Year Papers UP Polytechnic Previous Year Papers AP POLYCET Previous Year Papers TS POLYCET Previous Year Papers KEAM Previous Year Papers MHT CET Previous Year Papers WB JEE Previous Year Papers GUJCET Previous Year Papers ICAR AIEEA Previous Year Papers CUET PG Previous Year Papers JCECE Previous Year Papers Karnataka PGCET Previous Year Papers NEST Previous Year Papers KCET Previous Year Papers LPUNEST Previous Year Papers AMUEEE Previous Year Papers IISER IAT Previous Year Papers Bihar Diploma DECE-LE Previous Year Papers NPAT Previous Year Papers JMI Entrance Exam Previous Year Papers PGDBA Exam Previous Year Papers AP ECET Previous Year Papers PU CET Previous Year Papers GPAT Previous Year Papers CEED Previous Year Papers AIAPGET Previous Year Papers JKCET Previous Year Papers HPCET Previous Year Papers CG PAT Previous Year Papers SRMJEEE Previous Year Papers BCECE Previous Year Papers AGRICET Previous Year Papers TS PGECET Previous Year Papers MP PAT Previous Year Papers IIT JAM Previous Year Papers CMC Vellore Previous Year Papers ACET Previous Year Papers TS EAMCET Previous Year Papers NATA Previous Year Papers AIIMS MBBS Previous Year Papers BITSAT Previous Year Papers JEXPO Previous Year Papers HITSEEE Previous Year Papers AP EAPCET Previous Year Papers UCEED Previous Year Papers CG PET Previous Year Papers OUAT Previous Year Papers VITEEE Previous Year Papers
Syllabus
JEE Main Syllabus JEE Advanced Syllabus NEET Syllabus CUET Syllabus COMEDK UGET Syllabus UP Polytechnic JEECUP Syllabus AP POLYCET Syllabus TS POLYCET Syllabus KEAM Syllabus MHT CET Syllabus WB JEE Syllabus OJEE Syllabus ICAR AIEEA Syllabus CUET PG Syllabus NID Syllabus JCECE Syllabus Karnataka PGCET Syllabus NEST Syllabus KCET Syllabus UPESEAT EXAM Syllabus LPUNEST Syllabus PUBDET Syllabus AMUEEE Syllabus IISER IAT Syllabus NPAT Syllabus JIPMER Syllabus JMI Entrance Exam Syllabus AAU VET Syllabus PGDBA Exam Syllabus AP ECET Syllabus GCET Syllabus CEPT Syllabus PU CET Syllabus GPAT Syllabus CEED Syllabus AIAPGET Syllabus JKCET Syllabus HPCET Syllabus CG PAT Syllabus BCECE Syllabus AGRICET Syllabus TS PGECET Syllabus BEEE Syllabus MP PAT Syllabus MCAER PG CET Syllabus VITMEE Syllabus IIT JAM Syllabus CMC Vellore Syllabus AIMA UGAT Syllabus AIEED Syllabus ACET Syllabus TS EAMCET Syllabus PGIMER Exam Syllabus NATA Syllabus AFMC Syllabus AIIMS MBBS Syllabus BITSAT Syllabus BVP CET Syllabus JEXPO Syllabus HITSEEE Syllabus AP EAPCET Syllabus GITAM GAT Syllabus UPCATET Syllabus UCEED Syllabus CG PET Syllabus OUAT Syllabus IEMJEE Syllabus VITEEE Syllabus SEED Syllabus MU OET Syllabus
Books
Cut Off
JEE Main Cut Off JEE Advanced Cut Off NEET Cut Off CUET Cut Off COMEDK UGET Cut Off UP Polytechnic JEECUP Cut Off AP POLYCET Cut Off TNEA Cut Off TS POLYCET Cut Off KEAM Cut Off MHT CET Cut Off WB JEE Cut Off ICAR AIEEA Cut Off CUET PG Cut Off NID Cut Off JCECE Cut Off Karnataka PGCET Cut Off NEST Cut Off KCET Cut Off UPESEAT EXAM Cut Off AMUEEE Cut Off IISER IAT Cut Off Bihar Diploma DECE-LE Cut Off JIPMER Cut Off JMI Entrance Exam Cut Off PGDBA Exam Cut Off AP ECET Cut Off GCET Cut Off CEPT Cut Off PU CET Cut Off CEED Cut Off AIAPGET Cut Off JKCET Cut Off HPCET Cut Off CG PAT Cut Off SRMJEEE Cut Off TS PGECET Cut Off BEEE Cut Off MP PAT Cut Off VITMEE Cut Off IIT JAM Cut Off CMC Vellore Cut Off ACET Cut Off TS EAMCET Cut Off PGIMER Exam Cut Off NATA Cut Off AFMC Cut Off AIIMS MBBS Cut Off BITSAT Cut Off BVP CET Cut Off JEXPO Cut Off HITSEEE Cut Off AP EAPCET Cut Off GITAM GAT Cut Off UCEED Cut Off CG PET Cut Off OUAT Cut Off VITEEE Cut Off MU OET Cut Off
Latest Updates
Eligibility
JEE Main Eligibility JEE Advanced Eligibility NEET Eligibility CUET Eligibility COMEDK UGET Eligibility UP Polytechnic JEECUP Eligibility TNEA Eligibility TS POLYCET Eligibility KEAM Eligibility MHT CET Eligibility WB JEE Eligibility OJEE Eligibility ICAR AIEEA Eligibility CUET PG Eligibility NID Eligibility JCECE Eligibility Karnataka PGCET Eligibility NEST Eligibility KCET Eligibility LPUNEST Eligibility PUBDET Eligibility AMUEEE Eligibility IISER IAT Eligibility Bihar Diploma DECE-LE Eligibility NPAT Eligibility JIPMER Eligibility JMI Entrance Exam Eligibility AAU VET Eligibility PGDBA Exam Eligibility AP ECET Eligibility GCET Eligibility CEPT Eligibility PU CET Eligibility GPAT Eligibility CEED Eligibility AIAPGET Eligibility JKCET Eligibility HPCET Eligibility CG PAT Eligibility SRMJEEE Eligibility BCECE Eligibility AGRICET Eligibility TS PGECET Eligibility MP PAT Eligibility MCAER PG CET Eligibility VITMEE Eligibility IIT JAM Eligibility CMC Vellore Eligibility AIMA UGAT Eligibility AIEED Eligibility ACET Eligibility PGIMER Exam Eligibility CENTAC Eligibility NATA Eligibility AFMC Eligibility AIIMS MBBS Eligibility BITSAT Eligibility JEXPO Eligibility HITSEEE Eligibility AP EAPCET Eligibility GITAM GAT Eligibility UPCATET Eligibility UCEED Eligibility CG PET Eligibility OUAT Eligibility IEMJEE Eligibility SEED Eligibility MU OET Eligibility

Bisection Method: Definition, Steps & Solved Examples

Last Updated on Jul 01, 2025
Download As PDF
IMPORTANT LINKS

The bisection method is a simple way to find solutions to equations with only one unknown. It's a popular technique in mathematics for solving transcendental equations. 

In this mathematics article, we will delve into the bisection method and provide detailed explanations and examples to help you understand and apply it effectively.

Maths Notes Free PDFs

Topic PDF Link
Class 12 Maths Important Topics Free Notes PDF Download PDF
Class 10, 11 Mathematics Study Notes Download PDF
Most Asked Maths Questions in Exams Download PDF
Increasing and Decreasing Function in Maths Download PDF

What is the Bisection Method?

The Bisection Method is a simple way to find the root (or solution) of an equation when it is hard to solve it directly. It is used to find where a function becomes zero — in other words, where it crosses the x-axis. To use this method, we start with two numbers, a and b, such that the function values at these points have opposite signs (one positive and one negative). This tells us there is a root between them.

We then find the middle point of the interval and check the value of the function there. Based on the sign at the midpoint, we decide whether to look in the left or right half of the interval next. This process is repeated again and again until we get close enough to the actual root.

The Bisection Method is slow but very reliable and always works when the function is continuous and the initial values are chosen correctly.


Bisection Method Definition

The Bisection Method is a simple way to find the solution (root) of a polynomial equation. It is used when we know the solution lies between two numbers, say a and b. This method works only if the function is continuous and the values at a and b have opposite signs — one positive and one negative. This means the function crosses the x-axis somewhere between a and b.

The method keeps dividing the interval into two equal parts. In each step, it checks the midpoint of the interval and sees if the function becomes zero there. If not, it looks at the sign of the function at the midpoint and decides whether to move to the left half or the right half. This process is repeated until we get very close to the actual root.

The Bisection Method is also called the interval halving method, binary search method, or dichotomy method. It is slow but very reliable and always works if the function meets the required conditions. It’s a great method when you want a safe and steady way to find where a function equals zero.

Bisection Method Explanation

The bisection method can be used to solve the equation for a real variable , given that is a continuous function defined on the interval and and have opposite signs.

The process involves dividing the interval into two equal parts by calculating the midpoint, denoted as , and evaluating the function at that position. If is not the root, there exist two potential outcomes.

  • and exhibit different signs and enclose a root within their range.
  • and display different signs and enclose a root within their range.

A sub-interval is selected as the new interval for the subsequent step, and this repetition continues until the interval reaches a suitable level of smallness.

If the signs of and differ, then replaces . When the signs of and differ, replaces . If equals zero, is considered the solution, and the process concludes.

Test Series
133.1k Students
NCERT XI-XII Physics Foundation Pack Mock Test
323 TOTAL TESTS | 3 Free Tests
  • 3 Live Test
  • 163 Class XI Chapter Tests
  • 157 Class XII Chapter Tests

Get Started

How it works:

You start with two points, say a and b, where f(a) and f(b) have opposite signs. You then find the midpoint m = (a + b)/2 and check the value of f(m).

If f(m) is zero, you’ve found the root.

If not, check whether the sign of f(m) is the same as f(a) or f(b).

Replace the endpoint that has the same sign as f(m) with m and repeat the process.

This method keeps narrowing down the interval until the value is very close to the actual root. It’s slow but always gives the correct result if the function is continuous and changes sign in the interval.

The Bisection Method is widely used in mathematics and engineering for solving equations because it is easy to understand and apply.

Bisection Method Algorithm

To find the solution for a continuous function, follow these steps:

Consider a continuous function ,

Step 1: Find two points, and , where is smaller than , and the product of and is negative.

Step 2: Calculate the midpoint, , between and .

Step 3: If equals , then is the root of the function. If not, proceed to the next step.

Step 4: Divide the interval :

  • If the product of and is negative, there exists a root between and .
  • If the product of and is negative, there exists a root between and .

Step 5: Repeat the above three steps until equals .

In the above figure, the initial four stages of the bisection method involve taking certain steps. The presence of a significant blue dot denotes the root that has been discovered through four iterations. To achieve a higher level of accuracy, one can continue to repeat the aforementioned steps, adjusting the level of precision based on specific needs.

The bisection method is an approximation technique used to find the roots of an equation by repeatedly splitting the interval. This method keeps dividing the interval until a very narrow interval is obtained.

Advantages and Limitations of the Bisection Method

The Bisection Method offers several advantages:

  • Guaranteed convergence: Unlike some other methods, the Bisection Method guarantees convergence to a root within the initial interval.
  • Simplicity and ease of implementation: The method's straightforward nature makes it accessible to both beginners and experienced practitioners.
  • Robustness: The Bisection Method can handle various types of equations, including nonlinear and transcendental functions.

However, the method also has limitations:

  • Dependence on an initial interval: Selecting a suitable interval containing a root is crucial for successful implementation.
  • Slow convergence: The Bisection Method converges at a linear rate, which can be slower compared to methods like Newton-Raphson.
  • Inability to handle multiple roots: If an equation has multiple roots within the same interval, the Bisection Method may only find one of them.

Application of Bisection Method

The Bisection Method finds applications in diverse fields, including:

  • Engineering and physics: Solving equations related to mechanics, fluid dynamics, and electrical circuits.
  • Financial and economic modeling: Computing interest rates, asset pricing, and optimization problems.
  • Biological and ecological systems: Analyzing population dynamics, ecological modeling, and biochemical reactions.

Tips and Tricks on Bisection Method

Here are some tips and tricks for efficient implementation of the bisection method, presented in concise steps:

  • Choose an initial interval wisely to minimize the number of iterations.
  • Handle floating-point limitations by accounting for precision errors in calculations.
  • Implement a stopping criterion based on the desired level of accuracy.
  • Take advantage of function properties, such as monotonicity, to refine the interval selection.
  • Explore modifications and extensions of the method, such as the False Position Method, to improve convergence speed.

Properties of Bisection Method
  1. Bracketing Method
    The Bisection Method starts with two values, a and b, such that f(a) and f(b) have opposite signs. This means the function must cross the x-axis between a and b, so there is at least one root in that interval.
  2. Continuous Function Required
    The function f(x) must be continuous on the interval [a, b] for the method to work properly.
  3. Halves the Interval Each Time
    In every step, the method finds the midpoint of the interval and checks in which half the root lies. This process is repeated, making the interval smaller each time.
  4. Slow but Reliable
    The Bisection Method is slower compared to other methods (like Newton-Raphson), but it is very reliable and always converges if the initial conditions are correct.
  5. Guaranteed Convergence
    The method always converges to a root as long as the initial interval is correct (i.e., the signs of f(a) and f(b) are different).
  6. Accuracy Depends on Number of Iterations
    The more iterations you perform, the closer you get to the actual root. The error reduces by half in each step.

Bisection Method Solved Examples

Example 1: Find the root of the equation
f(x) = x³ - 2x - 5 within the interval [2, 3] using the Bisection Method.

Step 1: Evaluate f(2) and f(3)
f(2) = 2³ - 2×2 - 5 = 8 - 4 - 5 = -1
f(3) = 3³ - 2×3 - 5 = 27 - 6 - 5 = 16

Since f(2) < 0 and f(3) > 0, the function changes sign in the interval [2, 3], so a root exists.

Step 2: Apply the Bisection Method

Iteration 1:
Midpoint c = (2 + 3)/2 = 2.5
f(2.5) = (2.5)³ - 2×2.5 - 5 = 15.625 - 5 - 5 = 5.625
Root lies in [2, 2.5] (because f(2) < 0, f(2.5) > 0)

Iteration 2:
c = (2 + 2.5)/2 = 2.25
f(2.25) = (2.25)³ - 2×2.25 - 5 = 11.39 - 4.5 - 5 = 1.89
Root lies in [2, 2.25]

Iteration 3:
c = (2 + 2.25)/2 = 2.125
f(2.125) = 9.545 - 4.25 - 5 = 0.295

Iteration 4:
c = (2 + 2.125)/2 = 2.0625
f(2.0625) = 8.785 - 4.125 - 5 = -0.34
Root lies in [2.0625, 2.125]

Continue this process...

After several steps, we narrow the interval to around [2.09375, 2.095]
At this point, the root is approximately x ≈ 2.094 (correct to three decimal places).

2) Let's solve the equation within the interval .

Solution

Step 1: Evaluate the function at the interval endpoints:

First, we need to check if there's a root within the given interval. We can do this by evaluating f(x) at the endpoints:

  • f(-2) = e(-2) - (-2)3 - 5 = e(-2) + 8 - 5 = e(-2) + 3 = 0.135 + 3 = 3.135
  • f(-1) = e(-1) - (-1)3 - 5 = e(-1) + 1 - 5 = e(-1) - 4 = 0.368 - 4 = - 3.632

Step 2: Determine the new interval:

Since f(-2) is positive and f(-1) is negative, and f(x) is a continuous function, the Intermediate Value Theorem guarantees that there is at least one root within the interval [-2, -1].

So, we can bisect the interval and choose the right half for the next iteration:[-1.5, 1]

Repeat steps 1- 5 until the desired level of accuracy is reached. 

Let's assume we obtain a root within a tolerance of 0.001.

By following the steps, after several iterations, we find that the root of the equation 

f(x) = ex − x3 − 5 

within the interval [−2,−1] is approximately -1.689

We hope that the above article is helpful for your understanding and exam preparations. Stay tuned to the Testbook App for more updates on related topics from Mathematics and various such subjects. Also, reach out to the test series available to examine your knowledge regarding several exams.

Important Links
NEET Exam
NEET Previous Year Question Papers NEET Mock Test NEET Syllabus
CUET Exam
CUET Previous Year Question Papers CUET Mock Test CUET Syllabus
JEE Main Exam
JEE Main Previous Year Question Papers JEE Main Mock Test JEE Main Syllabus
JEE Advanced Exam
JEE Advanced Previous Year Question Papers JEE Advanced Mock Test JEE Advanced Syllabus

More Articles for Maths

FAQs For Bisection Method

The bisection method is a numerical root-finding technique that repeatedly divides an interval in half to narrow down and approximate the location of a root in an equation.

The method narrows down the range of possible solutions by evaluating the function at the endpoints of the interval and choosing the subinterval with opposite signs.

The bisection method is also known as the Interval Halving Method

The bisection method guarantees convergence to a root, is easy to implement, and can handle various types of equations.

The method depends on selecting a suitable initial interval, converges relatively slowly compared to other methods, and may find only one root within an interval.

The bisection method finds applications in fields such as engineering, physics, finance, economics, biology, and ecology for solving equations in diverse systems and models

It's called "bisection" because each time, the method cuts the interval into two equal parts to find a smaller range where the root exists.

Report An Error