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Question 1: a) Show that the equation x^2-23x+1=0 can be arranged to give x= {\Large \frac{x^2+1}{23}} b) Starting with x_0=0 use the iterative formula x_{n+1} = {\Large \frac{x_n^2+1}{23}} to obtain a solution of the equation x^2-23x+1=0 correct to 2 decimal places |

Question 2: a) Show that the equation x^2-13x+9=0 can be arranged to give x=- {\Large \frac{9}{x-13}} b) Starting with x_0=1 use the iterative formula x_{n+1} =- {\Large \frac{9}{x_n -13}} to obtain a solution of the equation x^2-13x+9=0 correct to 2 decimal places |

Question 3: a) Show that the equation x^2-19x+17=0 can be arranged to give x= \sqrt{19x-17} b) Starting with x_0=18 use the iterative formula x_{n+1} = \sqrt{19x_n -17} to obtain a solution of the equation x^2-19x+17=0 correct to 2 decimal places |

Question 4: a) Show that the equation x^2-27x+19=0 can be arranged to give x= {\Large \frac{x^2+19}{27}} b) Starting with x_0=1 use the iterative formula x_{n+1} = {\Large \frac{x_n^2+19}{27}} to obtain a solution of the equation x^2-27x+19=0 correct to 2 decimal places |

Question 5: a) Show that the equation x^2-25x+3=0 can be arranged to give x=- {\Large \frac{3}{x-25}} b) Starting with x_0=0 use the iterative formula x_{n+1} =- {\Large \frac{3}{x_n -25}} to obtain a solution of the equation x^2-25x+3=0 correct to 2 decimal places |

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Question 1: a) x^2-23x+1=0 \implies x^2+1 =23x \implies x= {\Large \frac {x^2+1}{23}} b) Root to 2 decimal places = 0.04 |

Question 2: a) x^2-13x+9=0 \implies x^2-13x=-9 \implies x(x-13)=-9 \implies x=- {\Large \frac{9}{x-13}} b) Root to 2 decimal places = 0.73 |

Question 3: a) x^2-19x+17=0 \implies x^2 =19x -17 \implies x= \sqrt{19x-17} b) Root to 2 decimal places = 18.06 |

Question 4: a) x^2-27x+19=0 \implies x^2+19 =27x \implies x= {\Large \frac {x^2+19}{27}} b) Root to 2 decimal places = 0.72 |

Question 5: a) x^2-25x+3=0 \implies x^2-25x=-3 \implies x(x-25)=-3 \implies x=- {\Large \frac{3}{x-25}} b) Root to 2 decimal places = 0.12 |

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