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Maths

Trial and improvement

This method involves substituting the unknown with different values, until we find one that works.

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 Rounding and estimating.

Example

Find the answer to the equation x3 – 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

  • 33 - (2 x 3)
  • 3 x 3 x 3 - (2 x 3)
  • 27 - 6
  • = 21; too small

Second try: x = 4

  • 43 - (2 x 4)
  • 4 x 4 x 4 - (2 x 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

  • 3.23 - (2 x 3.2)
  • 3.2 x 3.2 x 3.2 - (2 x 3.2)
  • 32.768 - 6.4
  • = 26.368; too high

Fourth try: x = 3.15

  • 3.153 - (2 x 3.15)
  • 3.15 x 3.15 x 3.15 - (2 x 3.15)
  • 31.255… - 6.3
  • = 24.955…; close

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.

Question

Solve the equation y2 + 2y = 40, correct to 1 decimal place.

toggle answer

Answer

Here is a worked solution:

y2 + 2y = 40

Let's start with y = 5

  • 5 × 5 + (2 × 5)
  • 25 + 10
  • = 35; too small

y = 6

  • 6 × 6 + (2 × 6)
  • 36 + 12
  • = 48; too big

So the answer lies between 5 and 6.

y = 5.5

  • 5.5 × 5.5 + (2 × 5.5)
  • 30.25 + 11
  • = 41.25; too big

y = 5.4

  • 5.4 5.4 + (2 × 5.4)
  • 29.16 + 10.8
  • = 39.96; too small

So the answer lies between 5.4 and 5.5.

y = 5.45

  • 5.45 × 5.45 + (2 × 5.45)
  • 29.7025 + 10.9
  • = 40.6025; too big

So the solution is between 5.4 and 5.5, and is less than 5.45. The question asks for an answer to 1 decimal place, so we round it down to 5.4.

The answer is: y = 5.4

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