Trial and Improvement

1. Trial and Improvement

2. We can solve this equation by: X2 + 2 = 11 X = 3 X2 = 9 Check:X2 + 2 = 3 x 3 +2 = 11

3. Look at this equation: X2 + 2 = 15 We can solve this in a similar way but for this type we are often asked to use trial and improvement This is a method where we substitute different values for x until we get as close as we can to the answer, usually one decimal place

4. Using trial and improvement show that the equation X2 + 2 = 15 has a solution between 3 and 4 The first thing we do is set up a table of values

5. In this column we write our guess or estimate Is the answer too big or too small? Write the equation here

6. Try a value half- way between 3 and 4 3 11 Too small Too big 4 18 Always write down all the digits on your calculator 3.5 14.25 Too small Too small 3.6 14.96 To decide if the solution is closer to 3.6 or 3.7, try the middle value 3.7 15.69 The solution is between 3.6 and 3.7 we cannot just guess which is closer Too big 3.65 Too big 3.65 15.3225

7. We only need the answer correct to one decimal place • X = 3.65 was too big. • So your next guess would be smaller • So you could try 3.64 or 3.63 • All these are 3.6 correct to 1 decimal place 3.65 3.6 3.7 So correct to one decimal place x would equal 3.6

8. Using trial and improvement show that the equation X3 + 2x = 7 has a solution between 1 and 2 The first thing we do is set up a table of values

9. In this column we write our guess or estimate Is the answer too big or too small? Write the equation here

10. Try the value half- way between 1 and 2 1 3 Too small Too big 2 12 1.5 To decide if the solution is closer to 1.5 or 1.6, try the middle value 6.375 Too small 7.296 Too big 1.6 1.55 1.55 6.823875 Too small The solution is between 1.5 and 1.6 remember we cannot just guess which is closer

11. We only need the answer correct to one decimal place • X = 1.55 was too small. • So your next guess would be bigger • You could try 1.56 or 1.57 • All these would be 1.6 correct to 1 decimal place 1.55 1.5 1.6 So correct to one decimal place x would equal 1.6

12. Using trial and improvement show that the equation X3 + x = 105 has a solution between 4 and 5 Set up a table of values

14. 4 68 Too small The solution is between 4.6 and 4.7. Remember, we cannot just guess which is closer Too big 5 130 To decide if the solution is closer to 4.6 or 4.7, try the middle value 4.5 Try the value half- way between 4 and 5 95.625 Too small Too small 4.6 101.936 4.7 108.523 Too big Too big 4.65 4.65 105.194625

15. We only need the answer correct to one decimal place • X = 4.65 was too big. • So your next guess would be smaller • You could try 4.64 or 4.63 • All these would be 4.6 correct to 1 decimal place 4.65 4.6 4.7 So correct to one decimal place x would equal 4.6

16. Using trial and improvement show that the equation 3X2 - 4x = 18 has a solution between 3 and 4 Set up your table of values

18. 3 15 Too small To decide if the solution is closer to 3.2 or 3.3, try the middle value Too big Try the value half- way between 3 and 4 4 32 3.5 22.75 Too big 21.08 Too big 3.4 3.3 19.47 Too big Remember to write down all the digits on your calculator 3.2 17.92 Too small The solution is between 3.2 and 3.3 remember, we cannot just guess which is closer 3.25 3.25 Too big 18.6875

19. We only need the answer correct to one decimal place • X = 3.25 was too big. • So your next guess would be smaller • So you could try 3.24 or 3.23 • All these would be 3.2 correct to 1 decimal place 3.25 3.2 3.3 So correct to one decimal place x would equal 3.2