### What is Bisection Method?

In mathematics, the bisection method in numerical techniques is used to find the actual root of a function. That applies to any function for finding the root.

To find the root we have to know about two values and their sign which is (positive)+ or -(negative).

The method repeatedly bisects the interval defined by these values and then selects the point at which the function changes the sign. It is a very simple and easy method but relatively slow. Because of this, it is used to obtain an approximation to a solution.

The bisection method is also called bracketing because it brackets the root within the interval.

#### The Method Of Bisection Rules

- The Bisection method is used for numerically solving the equation f(x) = 0 for the real variable x, where f is a continuous function defined on an interval [a, b].
- At each step, the method divides the interval point into two parts by computing the midpoint c = (a+b)/2 of the interval.

- After that, we put the c value at the function f(x).
- if the f(c) value is a positive value then the (b) value is replaced by the (c) value and if the f(c) value is negative then the (a) value is replaced by (c).

- if the f(c) value is a positive value then the f(b) value is replaced by the f(c) value and if the f(c) value is negative then the f(a) value is replaced by f(c).

**Bisection Method Algorithm**

- There are steps to find the root of a continuous function.
- For any continuous function f(x),
- Find two points a and b as an approximation root by applying values to the function.
- Find the midpoint c=a+b/2. Midpoint says c.
- C is the root of the given function if f(c) = 0;
- else follow the next step.
- Divide the interval [a, b] â€“ If f(c)*f(a) <0,
- there exist a root between c and aâ€“ else if f(c) *f (b) < 0, there exist a root between c and b
- Repeat above three steps until f(c) = 0.

### Example:

### Finding the root of a polynomial, the approximate root of f(x) = x^{3} + 5x + 1=0 , correct upto 3 decimal place.

Let the function is continous

F(x) = x^{3} + 5x + 1

Let put x=0 in f(x)

Then f(x)=1

Let put x=1 in f(x)

Then f(x)=7

Let put x=-1 in f(x)

Then f(x)=-5

Then us asign a value as (-1) and b value as (o)

It is the basic root

Mind the mid point c= a+b/c

C = -0.5

The find F(c)= x^{3} + 5x + 1

F(c) = -1.62

The f(c) value is negative so we replace the a value with c.

The steps are repeated up to correct 3 decimal points.

n | a | b | C | F(a) | F(b) | F(c) |

0 | -1 | 0 | -0.5 | -3 | 1 | -1.625 |

1 | -0.5 | 0 | -0.25 | -1.625 | 1 | -0.265 |

2 | -0.25 | 0 | -0.125 | -0.265 | 1 | 0.373 |

3 | -0.25 | -0.125 | -0.1875 | -0.265 | 0.373 | 0.0559 |

4 | -0.25 | -0.1875 | -0.21875 | -0.265 | 0.0559 | -0.1039 |

5 | -0.2187 | -0.1875 | -0.2031 | -0.1039 | 0.0559 | -0.0238 |

6 | -0.2031 | -0.1875 | -0.1953 | -0.0238 | 0.0559 | 0.0160 |

7 | -0.2031 | -0.1953 | -0.1992 | -0.0238 | 0.0160 | -3.9043 |

8 | -0.1992 | -0.1953 | -0.1972 | -3.9043 | 0.0160 | 6.0754 |

9 | -0.1992 | -0.1972 | -0.1982 | -3.9043 | 6.0754 | 1.2140 |

10 | -0.1992 | -0.1982 | -0.1987 | -3.9043 | 1.2140 | 0.0459 |

11 | -0.1992 | -0.1987 | -0.1989 | -3.9043 | 0.0459 | -2.6246 |

From the above calculation, we come to a conclusion that the reading c9 c10 c11 agree with one another for 3 decimal places is -0.198

Hence the approximation root of the function f(x)=**x ^{3} + 5x + 1** is -0.198 correct upto 3 decimal place by using bisection method.