Some cubic equations cannot be factorised easily, so approximate solutions are found using trial and improvement. This method tests values to find where the cubic expression changes sign and then improves the estimate until the solution is accurate enough.

 

A solution found by trial and improvement should always be supported by evidence that it lies between two values and the required accuracy should be justified.

 

 

Understanding the Method

To solve a cubic equation of the form \( f(x) = 0 \), trial and improvement involves:

1. Choose a value of \( x \) and calculate \( f(x) \)
2. Choose another value of \( x \) and calculate \( f(x) \)
3. Look for a change in sign between the results
4. Narrow the interval by testing values between them
5. A change in sign shows the solution lies between the two values.

 

 

Finding an Interval Containing a Solution

Consider:

$$
f(x) = x^3 - 5x - 1
$$

 

Test \( x = 2 \):

$$
f(2) = 2^3 - 5(2) - 1
$$

$$
= 8 - 10 - 1
$$

$$
= -3
$$

 

Test \( x = 3 \):

$$
f(3) = 3^3 - 5(3) - 1
$$

$$
= 27 - 15 - 1
$$

$$
= 11
$$

 

Since \( f(2) \) is negative and \( f(3) \) is positive, the solution lies between \( 2 \) and \( 3 \). 

 

 

Improving the Solution

Now test a value in between, such as \( 2.5 \):

$$
f(2.5) = 2.5^3 - 5(2.5) - 1
$$

$$
= 15.625 - 12.5 - 1
$$

$$
= 2.125
$$

 

This is positive, so the solution lies between \( 2 \) and \( 2.5 \).

 

Test \( 2.4 \):

$$
f(2.4) = 2.4^3 - 5(2.4) - 1
$$

$$
= 13.824 - 12 - 1
$$

$$
= 0.824
$$

 

Still positive, so the solution lies between \( 2 \) and \( 2.4 \).

 

Test \( 2.3 \):

$$
f(2.3) = 2.3^3 - 5(2.3) - 1
$$

$$
= 12.167 - 11.5 - 1
$$

$$
= -0.333
$$

 

Now the sign changes between \( 2.3 \) and \( 2.4 \), so the solution lies in that interval.

 

 

Justifying the Accuracy

If the question asks for the solution correct to one decimal place, the values \( 2.3 \) and \( 2.4 \) already justify this because the root lies between them, so the solution to 1 decimal place is \( 2.3 \) or \( 2.4 \) depending on which side the true root is closer to.

 

To justify correct to two decimal places, test \( 2.33 \) and \( 2.34 \) and show the sign change occurs within a smaller interval.

 

For example:

$$
f(2.33) = 2.33^3 - 5(2.33) - 1
$$

$$
= 12.649 - 11.65 - 1
$$

$$
= -0.001
$$

$$
f(2.34) = 2.34^3 - 5(2.34) - 1
$$

$$
= 12.813 - 11.7 - 1
$$

$$
= 0.113
$$

 

This shows the solution lies between \( 2.33 \) and \( 2.34 \). The width of this interval is \( 0.01 \), so the solution is correct to two decimal places.

 

 

Key Points to Remember

Trial and improvement finds approximate roots by checking values.
A change in sign shows the solution lies between two values.
Improve accuracy by testing values closer together.
Justify accuracy by showing the root lies within an interval of the required size.

 

Trial and improvement is a reliable method for solving cubic equations when exact algebraic solutions are not required or not practical.