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Worked Example
The following is an illustration of how equations may be solved by means of trial and improvement.
Solve x2 + 3x = 9
In this method, we try values of x in the equation, for example x = 2.
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22 + 3 x 2 = 10 |
(This would be too large) |
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| Next we could try x = 1 |
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12 + 3 x 1 = 4 |
(This would be too small) |
Then we could try a value half-way between 1 and 2 such as x = 1.5
Using a calculator, 1.52 + 3 x 1.5 = 6.75
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(This is too small)
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Try x = 1.8
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1.82 + 3 x 1.8 = 8.64
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(This is too small) |
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Try x = 1.9
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1.92 + 3 x 1.9 = 9.31
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(Too large) |
If we are asked for an answer to one decimal place we must try x = 1.85
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1.852 + 3 x 1.85 = 8.9725
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(Too small) |
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So x = 1.9
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(to one decimal place) |
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