Trial and improvement

When using trial and improvement, guess what the answer might be, then improve on it until you get close to the correct answer.

Example

A square has an area of {20}~cm^{2}. Use a method of trial and improvement to find the length of its side, correct to {1} decimal place.

  • Try length 4~cm

4 \times 4 = 16 (too small)

  • Try length 5~cm

5 \times 5 = 25 (too big)

So, the length must lie between 4~cm and 5~cm.

  • Try length 4.5~cm

4.5 \times 4.5 = 20.25 (too big)

  • Try length 4.4~cm

4.4 \times 4.4 = 19.36 (too small)

Therefore, the answer lies between 4.4 and 4.5.

Now, we need to test halfway between these two values.

{4.45}\times{4.45}={19.8025}

As this value is less than 20, the answer is closer to 4.5.

Because the answer lies between 4.4 and 4.5, and is closer to 4.5, we can say that the length of the side of the square is 4.5~cm correct to 1 decimal place.

More accurate answers

The question asked us to find the length of the side correct to 1 decimal place, so initially you need to try values with 1 decimal place. You would then have to test halfway between two adjacent values to decide on the final answer.

If, however, the question had asked for values correct to 2 decimal places, you would have got a more accurate answer. For example:

4.45 \times 4.45 = 19.80 (too small).

We already know that the length must lie between 4.4~cm and 4.5~cm.

  • Try length 4.45~cm

4.45 \times 4.45 = 19.8025 (too small)

  • Try length 4.46~cm

4.46 \times 4.46 = 19.8916 (too small)

  • Try length 4.47~cm

4.47 \times 4.47 = 19.9809 (too small)

  • Try length 4.48~cm

4.48 \times 4.48 = 20.0704 (too big)

Therefore, the answer lies between 4.47 and 4.48

Now, we need to test half way between these two values.

4.475 \times 4.475 = 20.025625

As this value is more than 20, the answer is closer to 4.47.

The length of the side of the square is 4.47~cm correct to 2 decimal places.

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