Trial and Improvement

1. Trial and Improvement

2. Trial and Improvement x3 + x = 10 The answer for this equation falls between 3 and 4. Using trial and improvement, find the solution to 1 decimal place. First, draw a table:

3. Trial and Improvement x2 + x = 10 If you know that the answer is between 2 and 3, it makes sense to start at 2.5: 2.52 + 2.5 = 8.75 This is too small so we need to try a bigger number...

4. Trial and Improvement x2 + x = 10 2.5 was too small so we should try 2.6: 2.62 + 2.6 = 9.36 This is still too small so we need to try a bigger number...

5. Trial and Improvement x2 + x = 10 2.5 and 2.6 were too small so let’s try 2.8: 2.82 + 2.8 = 10.64 This is too big so we need to try a smaller number...

6. Trial and Improvement x2 + x = 10 2.6 was too small and 2.8 was too big so let’s try 2.7: 2.72 + 2.7 = 9.99 This is too big so we need to try a smaller number...

7. Trial and Improvement x2 + x = 10 As the answer has to be to 1 decimal place, which of the following result is closest to 10? x = 2.7 (1dp)