# Maths change of sign coursework

Extracts from this document...

Introduction

Osaamah Mohammed

## Maths A2 Coursework

Change of Sign Method:-

3x3 – 6x + 1 = 0

##### Sketch: -

##### Looking at the graph we find that the roots lie between these intervals:

##### [-2,-1] [0,1] [1,2]

Finding root interval between [0,1] using a Decimal Search

### Root [0 , 1]

X1 = 0

∴

Middle

0.17

-0.005261

### Root [0.16 , 0.17]

X | f(X) |

0.16 | 0.052298 |

0.161 | 0.046520 |

0.162 | 0.040755 |

0.163 | 0.034992 |

0.164 | 0.029233 |

0.165 | 0.023476 |

0.166 | 0.017723 |

0.167 | 0.011972 |

0.168 | 0.0062249 |

0.169 | 0.00048043 |

0.170 | -0.005261 |

#### Root [0.169 , 0.170]

X | f(X) |

0.1690 | 0.00048043 |

0.1691 | -0.0000923852 |

### Root [0.1690 , 0.1691]

X | f(X) |

0.1690 | 0.00048043 |

0.16901 | 0.00042299 |

0.16902 | 0.00036557 |

0.16903 | 0.00030814 |

0.16904 | 0.00025071 |

0.16905 | 0.00019328 |

0.16906 | 0.00013586 |

0.16907 | 0.00078427 |

0.16908 | 0.00021 |

0.16909 | -0.000036426 |

From this method we find that the root lies between[0.16908 , 0.16909]

∴ Root = 0.169085 ± 0.000005

testf (0.16908) = 0.00021

f (0.16909) = -0.000036426

There is a change

Conclusion

Start X1 = 0 Use Xn+1 = 2xn3 + 1

5

X2 = 2(0)3 + 1 = 0.2

5

X3 = 2(0.2)3 + 1 = 0.216

5

X4 = 2(0.216)3 + 1 = 0.22016

5

X5 = 0.22134

X6= 0.22169

X7 = 0.22179

X8 = 0.22182

X9 = 0.22183 root to 5 s.f (took 9 iterations)

X10 = 0.22183

Failure: -

Solve 2x3 – 5x + 1 = 0

Re-arrange this so that “X=….”

- 2x3 – 5x + 1 = 0

2x3 +1 = 5x

2x3 = 5x – 1

x3 = 5x – 1

2

x = (5x – 1)⅓

2

x = g(x)

Sketch : -

This student written piece of work is one of many that can be found in our AS and A Level Core & Pure Mathematics section.

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