Using bisection method, find the root of the equation f(x)=x^{2} -6logx-3=0(2<=x<=3)'correct to 4 decimal places with an accuracy of 10^{ -2} .

Here, f(x)=x^{2} -6 logx-3=0

f(2)=4-6 log2-3=-0.806

f(3)=9-6 log3-3=3.1373

f(2).f(3)=-0.806*3.1373=-2.529422 which is negative.

Hence, the root lies between 2 and 3

c_{0} =(2+3)/2=2.5

f(2.5)=6.25-6 log 2.5-3=0.8623

Now

n | a(-ve) | b(+ve) | c_{n} | f(c_{n}) |

0 | 2 | 3 | 2.5 | 0.8623 |

1 | 2 | 2.5 | 2.25 | -0.050595 |

2 | 2.25 | 2.5 | 2.375 | 0.38664 |

3 | 2.25 | 2.375 | 2.3125 | 0.1631658 |

4 | 2.25 | 2.3125 | 2.28125 | 0.05506 |

5 | 2.25 | 2.28125 | 2.265625 | 0.001925 |

From the table,

f(2.265625)=0.001928<10^{-2}

Therefore, the root of the given equation correct upto four paces of decimal=2.2656

Here, f(x)=x^{2} -6 logx-3=0

f(2)=4-6 log2-3=-0.806

f(3)=9-6 log3-3=3.1373

f(2).f(3)=-0.806*3.1373=-2.529422 which is negative.

Hence, the root lies between 2 and 3

c_{0} =(2+3)/2=2.5

f(2.5)=6.25-6 log 2.5-3=0.8623

Now

n | a(-ve) | b(+ve) | c_{n} | f(c_{n}) |

0 | 2 | 3 | 2.5 | 0.8623 |

1 | 2 | 2.5 | 2.25 | -0.050595 |

2 | 2.25 | 2.5 | 2.375 | 0.38664 |

3 | 2.25 | 2.375 | 2.3125 | 0.1631658 |

4 | 2.25 | 2.3125 | 2.28125 | 0.05506 |

5 | 2.25 | 2.28125 | 2.265625 | 0.001925 |

From the table,

f(2.265625)=0.001928<10^{-2}

Therefore, the root of the given equation correct upto four paces of decimal=2.2656