Trial and improvement

You might need to use this method if you are asked to solve an equation where there is no exact answer. You may also be asked to give the solution to a given number of decimal places or significant figures. The question should indicate the degree of accuracy required.

For a quick recap on rounding see our section Approximation.

Example

Find the answer to the equation x^3 – 2x = 25 to one decimal place.

Solution

We are looking for a number to replace x, that when applied to the equation will give us 25. Start by guessing what x could be, then refine your answer based on your result. Set it out like this.

First we'll try: x = 3

3^3 - (2 \times 3)

27 - 6

= 21; too small

Second try: x = 4

4^3 - (2 \times 4)

64 - 8

= 56; much too high

We could use a number half way between 3 and 4, but our first tries suggest it will be closer to 3.

Third try: x = 3.2

3.2^3 - (2 \times 3.2)

32.768 - 6.4

= 26.368; too high

Fourth try: x = 3.1

3.1^3 - (2 \times 3.1)

29.791 - 6.2

= 23.591; too small

This means the answer lies between 3.1 and 3.2.

Since we've been told to give the answer correct to 1 decimal place, we have to try x = 3.15 to determine if the answer is closer to 3.1 or 3.2.

Try: x = 3.15

3.15^3 - (2 \times 3.15)

3.15 \times 3.15 \times 3.15 - (2 \times 3.15)

31.255... - 6.3

= 24.955975

This means the actual value of x is greater than 3.15 but less than 3.2.

Since we've been told to give the answer correct to 1 decimal place, the answer we are looking for is 3.2.