## Example Question

a) Show that the equation x^{3}- 6x + 1 = 0 has a solution between 2 and 3.

b) Solve this equation correct to 1 decimal place.

## Solution

a) To show that x^{3}- 6x + 1 = 0 has a solution between 2 and 3 we need to substitute x = 2 and x = 3 into the equation.

When x = 2 we get: (2)

^{3}- 6(2) + 1 = 8 - 12 + 1 = -3.

When x = 3 we get: (3)

^{3}- 6(3) + 1 = 27 - 6(3) + 1 = 10.

Now notice when x = 2 we get an answer less than 0 and when x = 3 we get an answer greater than 0. This means there must be a value of x between 2 and 3 which is equal to 0.

b) We now need to find this solution correct to 1 decimal place. To do this we need to draw a table to test different values of x. Since we know the answer is between 2 and 3 it makes sense to start with x = 2.5.

x | x^{3} - 6x + 1 |
Comment |
---|---|---|

2 | -3 | Too small |

3 | 10 | Too big |

2.5 | 1.625 | Too big |

2.3 | -0.633 | Too small |

2.4 | 0.424 | Too big |

Now since 2.3 was too small and 2.4 was too big we know the solution is between these values. To find our answer to 1 decimal place we have to try one more value in the middle of these, when x = 2.35.

When x = 2.35 we get: (2.35)

^{3}- 6(2.35) + 1 = -0.122125.

Finally since 2.35 is too small then we can say 2.3 must also be too small. Hence x = 2.4 to 1 decimal place.

## Test Yourself!

The equation x^{3}+ x = 15 has a solution between 2 and 3.

Find this solution correct to one decimal place.

x =

The equation x

^{3}- 2x + 6 = 0 has a solution between -3 and -2.

Find this solution correct to one decimal place.

x =