# Bisection Method

In Mathematics, the bisection method is a straightforward technique to find numerical solutions of an equation with one unknown. Among all the numerical methods, the bisection method is the simplest one to solve the transcendental equation. In this article, we will discuss the bisection method with solved problems in detail.

## Bisection Method Definition

The bisection method is used to find the roots of a polynomial equation. It separates the interval and subdivides the interval in which the root of the equation lies. The principle behind this method is the intermediate theorem for continuous functions. It works by narrowing the gap between the positive and negative intervals until it closes in on the correct answer. This method narrows the gap by taking the average of the positive and negative intervals. It is a simple method and it is relatively slow. The bisection method is also known as interval halving method, root-finding method, binary search method or dichotomy method.

Let us consider a continuous function “f” which is defined on the closed interval [a, b], is given with f(a) and f(b) of different signs. Then by intermediate theorem, there exists a point x belong to (a, b) for which f(x) = 0.

## Bisection Method Algorithm

Follow the below procedure to get the solution for the continuous function:

For any continuous function f(x),

- Find two points, say a and b such that a < b and f(a)* f(b) < 0
- Find the midpoint of a and b, say “t”
- t is the root of the given function if f(t) = 0; else follow the next step
- Divide the interval [a, b]
– If f(t)*f(a) <0, there exist a root between t and a

– else if f(t) *f (b) < 0, there exist a root between t and b - Repeat above three steps until f(t) = 0.

The bisection method is an approximation method to find the roots of the given equation by repeatedly dividing the interval. This method will divide the interval until the resulting interval is found, which is extremely small.

### Bisection Method Example

**Question:** Determine the root of the given equation x^{2}-3 = 0 for x ∈ [1, 2]

**Solution:**

Given: x^{2}-3 = 0

Let f(x) = x^{2}-3

Now, find the value of f(x) at a= 1 and b=2.

f(x=1) = 1^{2}-3 = 1 – 3 = -2 < 0

f(x=2) = 2^{2}-3 = 4 – 3 = 1 > 0

The given function is continuous, and the root lies in the interval [1, 2].

Let “t” be the midpoint of the interval.

I.e., t = (1+2)/2

t =3 / 2

t = 1.5

Therefore, the value of the function at “t” is

f(t) = f(1.5) = (1.5)^{2}-3 = 2.25 – 3 = -0.75 < 0

If f(t)<0, assume a = t.

and

If f(t)>0, assume b = t.

f(t) is negative, so a is replaced with t = 1.5 for the next iterations.

The iterations for the given functions are:

Iterations |
a |
b |
t |
f(a) |
f(b) |
f(t) |

1 |
1 | 2 | 1.5 | -2 | 1 | -0.75 |

2 |
1.5 | 2 | 1.75 | -0.75 | 1 | 0.062 |

3 |
1.5 | 1.75 | 1.625 | -0.75 | 0.0625 | -0.359 |

4 |
1.625 | 1.75 | 1.6875 | -0.3594 | 0.0625 | -0.1523 |

5 |
1.6875 | 1.75 | 1.7188 | -01523 | 0.0625 | -0.0457 |

6 |
1.7188 | 1.75 | 1.7344 | -0.0457 | 0.0625 | 0.0081 |

7 |
1.7188 | 1.7344 | 1.7266 | -0.0457 | 0.0081 | -0.0189 |

So, at the seventh iteration, we get the final interval [1.7266, 1.7344]

Hence, 1.7344 is the approximated solution.

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