Show that the equation \(x^3 - 6x^2 + 12 = 0\) has two positive roots and one negative root.

Let \(f(x) = x^3 - 6x^2 + 12\), which is a cubic, and thus has at most three real roots. We can tabulate various values for \(f(x)\).

\(x\) | \(f(x)\) |
---|---|

-2 | -20 |

-1 | 5 |

0 | 12 |

1 | 7 |

2 | -4 |

3 | -15 |

4 | -20 |

5 | -13 |

6 | 12 |

The function \(f(x)\) changes sign three times in the table, so that \(f(x)\) has three real roots, one between \(x = -2\) and \(x = -1\); one between \(x = 1\) and \(x = 2\); and a third between \(x = 5\) and \(x = 6\).

So \(f(x)\) has two positive roots and one negative root.

Show that the negative root lies between \(-1\) and \(-1.5\)…

We saw above that \(f(-1) = 5\). As \(f(-1.5) = -\dfrac{39}{8} < 0\), the negative root must lie between \(x = -1\) and \(x = -1.5\).

…and find it correct to one decimal place.

We can tabulate more values for \(f(x)\).

\(x\) | \(f(x)\) |
---|---|

-1.5 | -4.875 |

-1.4 | -2.504 |

-1.3 | -0.337 |

-1.2 | 1.632 |

-1.1 | 3.409 |

-1 | 5.0 |

Thus the negative root lies between \(x = -1.3\) and \(x = -1.2\).

We have \(f(-1.25) = 0.671875 > 0\), and so the root falls between \(x = -1.3\) and \(x = -1.25\), so that, to one decimal place, the negative root is \(x = -1.3\).