1. Algebra Trial and Improvement We know how to solve an equation where the answer is an integer (1, 2, 4 etc) Sometimes we need to find an answer to an equation which is only approximate (close to 3.6). To do this we need to use the trial and improvement method.

2. Algebra Trial and Improvement Solve the equation x 3 + x = 105 to 1 decimal place Step 1 – chose any number that is likely to be in the right sort of region. (in this case 5 would be a good start). When we replace x with 5 the answer is: (5 x 5 x 5) + 5 = 125 + 5 = 130 (answer is too big)

3. Algebra Trial and Improvement Solve the equation x 3 + x = 105 Step 2 – replace x with 4: (4 x 4 x 4) + 4 = 64 + 4 = 68 (answer is too low. The answer must be somewhere between 4 and 5. Next time we need to use 4.5)

4. Algebra Trial and Improvement Solve the equation x 3 + x = 105 Step 2 – replace x with 4.5: (4.5 x 4.5 x 4.5) + 4.5 = 91.125 + 4.5 = 95.625 (answer is still too low but is much closer. We try increasing the number to 4.6)

5. Algebra Trial and Improvement Solve the equation x 3 + x = 105 Step 2 – replace x with 4.6: (4.6 x 4.6 x 4.6) + 4.6 = 97.336 + 4.6 = 101.936 (answer is still too low but is even closer. We try increasing the number to 4.7)

6. Algebra Trial and Improvement Solve the equation x 3 + x = 105 Step 2 – replace x with 4.7: (4.7 x 4.7 x 4.7) + 4.7 = 103.823 + 4.6 = 108.523 (answer is now too big. The answer lies somewhere between 4.6 and 4.7. As we have to find the answer to 1 decimal point only, we have to chose either 4.6 or 4.7. But which?)

7. Algebra Trial and Improvement Solve the equation x 3 + x = 105 Step 2 – replace x with 4.65, half way between 4.6 and 4.7: (4.65 x 4.65 x 4.65) + 4.65 = 100.544 + 4.65 = 105.194 (answer is now almost spot on but just a fraction over 105, the target number. So 4.6 is the nearest number. x = 4.6)