±«Óătv

Trial and improvement

Sometimes we must use trial and improvement to approximate the result of cubic equations.

Example

The solution to the equation \({x}^{3}~–~{15x~+~5}={2}\) lies between 3 and 4. Use trial and improvement to find a solution correct to one decimal point.

Solution

In order to tackle this question we must first construct a table:

A table with 2 rows and 3 columns labelled "x", "Outcome" and "Too big/small"

This is where we will record the steps we have taken in order to find the solution. We know the answer lies between 3 and 4 so we will start by trying 3.5 (as shown in the table).

As the outcome was too small, we will now try a larger number: 3.7

A table with 3 rows and 3 columns labelled "x", "Outcome" and "Too big/small"

The answer is still too small so we shall try 3.8

A table with 4 rows and 3 columns labelled "x", "Outcome" and "Too big/small"

Now the answer is too large, but we also know that 3.7 is too small. As we are asked to give our answer correct to one decimal place, we must either choose 3.7 or 3.8. The way we decide is by using 3.75. If this is too big, then our answer will be 3.7. If it is too small, our answer will be 3.8.

A completed table with 4 rows and 3 columns labelled "x", "Outcome" and "Too big/small"

As 3.75 was too small, our final answer is \({x}\) = 3.8 (correct to one decimal place).