## Some Questions in the Bisection method

There are some questions about the Bisection method in numerical techniques. before you read the questions you must know about the Bisection method. Previously we discuss the BiSection Method In Numerical Techniques. If you haven’t seen it then hurry up!

#### Questions No 1 :

**Find the approximation to √3 correct up to two decimal points using the bisection method.**

Let x = √3

Squaring both side

=> X^{2 }= (√3)^{2}

=> X^{2 }= 3

=> X^{2}-3 = 0

Let the function is continuous at** **

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

F(x) = X^{2}-3

· Let put x=0 in f(x)

Then f(x)= -3

· Let put x=1 in f(x)

Then f(x) = -2

· Let put x=2 in f(x)

Then f(x)= 1

· Let put x=-1 in f(x)

Then f(x) = -2

From the above calculation we take (a=1) and (b=2) as approxmate root

Mind the midpoint c= a + b /c

C = 1+2/2

=1.5

The find F(c) = X^{2}-3

F(c) = -0.75

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

The steps are repeated up to correct 2 decimal points.

The table is given below

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

0 | 1 | 2 | 1.5 | -2 | 1 | -0.75 |

1 | 1.5 | 2 | 1.75 | -0.75 | 1 | 0.0625 |

2 | 1.5 | 1.75 | 1.625 | -0.75 | 0.0625 | -0.3593 |

3 | 1.625 | 1.75 | 1.6875 | -0.3593 | 0.0625 | -0.1523 |

4 | 1.6875 | 1.75 | 1.7187 | -0.1523 | 0.0625 | -0.0458 |

5 | 1.7187 | 1.75 | 1.7343 | -0.0458 | 0.0625 | 7.7964 |

6 | 1.7187 | 1.7343 | 1.7265 | -0.0458 | 7.7964 | -0.0191 |

7 | 1.7265 | 1.7343 | 1.7304 | -0.0191 | 7.7964 | -5.7158 |

8 | 1.7304 | 1.7343 | 1.7323 | -5.7158 | 7.7964 | 8.6329 |

9 | 1.7304 | 1.7323 | 1.7313 | -5.7158 | 8.6329 | 8.6329 |

From the above calculation, we come to a conclusion that the reading c7 c8 c9 are agree with one another for 2 decimal places is 1.73

Hence the approximation root of function √3 is 1.73 correct up to 2 decimal places by using the bisection method.

#### Question No 2 :

### Find the approximation to √2 correct up to two decimal points using the bisection method.

Let x = √2

Squaring both side

=> X^{2 }= (√2)^{2}

=> X^{2 }= 2

=> X^{2}-2 = 0

Let the function is continuous at** **

f(x) = X^{2}-2

F(x) = X^{2}-2

· Let put x=0 in f(x)

Then f(x)= -2

· Let put x=1 in f(x)

Then f(x)= -1

· Let put x=2 in f(x)

Then f(x)= 2

· Let put x=-1 in f(x)

** ** Then f(x)= -1

From the above calculation we take (-1) and (2) as approx root

Mind the midpoint c= a + b /c

C = -1+2/2

=1.5

The find F(c) = X^{2}-3

F(c) = -0.75

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

The steps are repeated up to correct 2 decimal points.

The table is given below

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

0 | -1 | 2 | 0.5 | -1 | 2 | -1.75 |

1 | 0.5 | 2 | 1.25 | -1.75 | 2 | -0.4375 |

2 | 1.25 | 2 | 1.625 | -0.4375 | 2 | 0.6406 |

3 | 1.25 | 1.625 | 1.4375 | -0.4375 | 0.6406 | 0.0664 |

4 | 1.25 | 1.4375 | 1.3437 | -0.4375 | 0.0664 | -0.1944 |

5 | 1.3437 | 1.4375 | 1.3906 | -0.1944 | 0.0664 | -0.0662 |

6 | 1.3906 | 1.4375 | 1.4140 | -0.0662 | 0.0664 | -6.04 |

7 | 1.4140 | 1.4375 | 1.4257 | -6.04 | 0.0664 | 0.0326 |

8 | 1.4140 | 1.4257 | 1.4198 | -6.04 | 0.0326 | 0.0158 |

9 | 1.4140 | 1.4198 | 1.4169 | -6.04 | 0.0158 | 7.6056 |

10 | 1.4140 | 1.4169 | 1.4154 | -6.04 | 7.6056 | 3.3571 |

From the above calculation, we come to a conclusion that the reading c8 c9 c10 agree with one another for 2 decimal places is 1.41

Hence the approximation root of function √2 is 1.41 correct up to 2 decimal places by using the bisection method.

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