Dr Frost Year 8 Trial and Improvement Objectives

ζ Dr Frost Year 8 – Trial and Improvement Objectives: Be able to find approximate solutions to more difficult equations by gradually refining our answer.

Ranges on Number Lines We can use number lines to express a range of values that are possible. Means the value is included. ! Means the value is NOT included. As an inequality: ? On number line: ? 2 4 6 8 9 As an inequality: ? On number line: ? 3. 9 4. 0 4. 1 4. 2 4. 3

Starter ?

Trial and Improvement How did you know when to try bigger or smaller values of x on the next step? ! x = 5 x = 10 x = 8. 5 x = 8. 8 x = 8. 9 x = 8. 87 x = 8. 88 : 5(5 -2) = 15? : 10(10 -2) = 80 ? : 8(8 -2) = 48? : 8. 5(8. 5 -2) = 55. 25 ? : 8. 8(8. 8 -2) = 59. 84 ? : 8. 9(8. 9 -2) = 61. 41 ? : 8. 87(8. 87 -2) = 60. 94 ? : 8. 88(8. 88 -2) = 61. 09 ? Too small Too large Too small Too large

Showing there’s a solution ? ?

When to stop? Solve x(x-2) = 61 x = 8. 87 x = 8. 88 : 8. 87(8. 87 -2) = 60. 94 : 8. 88(8. 88 -2) = 61. 09 Too small Too large Suppose the we wanted the answer correct to 2 dp. Could we stop at this point? What value would we choose for x? x = 8. 875 : 8. 875(8. 875 -2) = 61. 02 Too large ? We know therefore that the value of x lies between: ? 8. 87 and 8. 875. ? To 2 dp the solution must be 8. 87.

Another Example A container in the shape of a cuboid with a square base is to be constructed. The height of the cuboid is to be 2 metres less than the length of a side of its base and the container is to have a volume of 45 cubic metres. ? ?

Exercises Edexcel GCSE Mathematics Page 25 B – Page 419 Q 1 a, 2 a, c, 4, 6, 8, 10

What have we learnt? We can find solutions to equations by gradually improving our estimate. Solve 2 x = 14 to 2 dp. . . x = 3. 80 : 23. 80 = 13. 93 Too small x = 3. 81 : 3. 32 + 3. 3 = 14. 03 Too large ? x = 3. 805 : 3. 252 + 3. 25 = 13. 98 Too small ? So x = 3. 81 to 2 dp ? Solve x 2 = 4 + x to 1 dp Solve x 2 – x = 4. . . x = 2. 5 : 2. 52 – 2. 5 = 3. 75 Too small x = 2. 6 : 2. 62 – 2. 6 = 4. 16 Too large x = 2. 55 : 2. 552 – 2. 55 = 3. 95 Too small ? So x = 2. 6 to 1 dp ?

Puzzle A square of side length 2 is cut of a circle of radius x. The resulting area is x. Form an equation involving the area, and hence use trial and improvement to determine x correct to 1 dp. Equation for area: πx 2 – 4 = x π x 2? – x = 4 x 2 2 x = 1. 2 : 3. 324 x = 1. 3 : 4. 009 x = 1. 25 : 3. 659 ? Too small Too large Too small So x = 1. 3 ? to 1 dp Area = x Work in 3 s/4 s (but you need to each individually show your working in your book)