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Year 8 – Trial and Improvement

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.

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Year 8 – Trial and Improvement

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

  2. 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. ! A weight being 6kg correct to the nearest 2kg As an inequality: On number line: ? ? 2 4 6 8 9 A length being 4.1 correct to 1dp On number line: As an inequality: ? ? 3.9 4.0 4.1 4.2 4.3

  3. Starter You want to solve the equation: Use ‘trial and error’ to find the most accurate value of that you can. = 8.874007874011... ?

  4. Trial and Improvement How did you know when to try bigger or smaller values of x on the next step? ! Solve x = 5 : 5(5-2) = 15 Too small x = 10 : 10(10-2) = 80 Too large x = 8 : 8(8-2) = 48 Too small x = 8.5 : 8.5(8.5-2) = 55.25 Too small x = 8.8 : 8.8(8.8-2) = 59.84 Too small x = 8.9 : 8.9(8.9-2) = 61.41 Too large x = 8.87 : 8.87(8.87-2) = 60.94 Too small x = 8.88 : 8.88(8.88-2) = 61.09 Too large ? ? ? ? ? ? ? ?

  5. Showing there’s a solution Show that has a solution between 1 and 2. When , . Since , this is too small. When , . Since this is too big. Thus solution must lie between 1 and 2. ? Show that has a solution between 1 and 3. ? When , . Too small. When , . Too big. Thus solution must lie between 1 and 3.

  6. When to stop? Solve x(x-2) = 61 x = 8.87 : 8.87(8.87-2) = 60.94 Too small x = 8.88 : 8.88(8.88-2) = 61.09 Too large Suppose the we wanted the answer correct to 2dp. 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 2dp the solution must be 8.87. ? ?

  7. 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. Taking metres as the length of a side of the base, show that satisfies the equation Use a trial and improvement method to find the solution of the equation that lies between 4 and 5. Give your answer correct to two decimal places. Find the height of the container correct to the nearest cm. ? • : Too small • : Too big • : Too big • So ?

  8. Exercises Edexcel GCSE Mathematics Page 25B – Page 419 Q1a, 2a, c, 4, 6, 8, 10

  9. What have we learnt? We can find solutions to equations by gradually improving our estimate. Solve 2x = 14 to 2dp ... 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 2dp ? ? Solve x2 = 4 + x to 1dp  Solve x2 – 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 1dp ? ? ?

  10. Puzzle A square of side length 2 is cut out 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 1dp. Equation for area: πx2 – 4 = x  π x2 – x = 4 ? x 2 x = 1.2 : 3.324 Too small x = 1.3 : 4.009 Too large x = 1.25 : 3.659 Too small So x = 1.3 to 1dp ? 2 ? Area = x Work in 3s/4s (but you need to each individually show your working in your book)

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