ζYear 8 – Trial and ImprovementDr Frost

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.

5≤𝑤<7 2 4 6 8 9

A length being 4.1 correct to 1dp

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3.9 4.0 4.1 4.2 4.3

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?4.05≤𝑥<4.15As an inequality:

On number line:

A weight being 6kg correct to the nearest 2kg

As an inequality: On number line:

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You want to solve the equation:

Use ‘trial and error’ to find the most accurate value of that you can.

= 8.874007874011...?

Starter

x = 5 : 5(5-2) = 15 Too smallx = 10 : 10(10-2) = 80 Too largex = 8 : 8(8-2) = 48 Too smallx = 8.5 : 8.5(8.5-2) = 55.25 Too smallx = 8.8 : 8.8(8.8-2) = 59.84 Too smallx = 8.9 : 8.9(8.9-2) = 61.41 Too largex = 8.87 : 8.87(8.87-2) = 60.94 Too smallx = 8.88 : 8.88(8.88-2) = 61.09 Too large

Trial and Improvement

How did you know when to try bigger or smaller values of x on the next step?

! Solve

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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.

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When to stop?

Suppose the we wanted the answer correct to 2dp.Could we stop at this point? What value would we choose for x?

Solve x(x-2) = 61

x = 8.875 : 8.875(8.875-2) = 61.02 Too large

x = 8.87 : 8.87(8.87-2) = 60.94 Too smallx = 8.88 : 8.88(8.88-2) = 61.09 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.

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: Too small: Too big: Too big

So

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.

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Another Example

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a) Taking metres as the length of a side of the base, show that satisfies the equation

b) 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.

c) Find the height of the container correct to the nearest cm.

Exercises

Edexcel GCSE Mathematics

Page 25B – Page 419Q1a, 2a, c, 4, 6, 8, 10

Solve 2x = 14 to 2dp...x = 3.80 : 23.80 = 13.93 Too smallx = 3.81 : 3.32 + 3.3 = 14.03 Too largex = 3.805 : 3.252 + 3.25 = 13.98 Too small

So x = 3.81 to 2dp

We can find solutions to equations by gradually improving our estimate.

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What have we learnt?

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Solve x2 = 4 + x to 1dp Solve x2 – x = 4...x = 2.5 : 2.52 – 2.5 = 3.75 Too smallx = 2.6 : 2.62 – 2.6 = 4.16 Too largex = 2.55 : 2.552 – 2.55 = 3.95 Too small

So x = 2.6 to 1dp?

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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.

2

2

x

Area = x

x = 1.2 : 3.324 Too smallx = 1.3 : 4.009 Too largex = 1.25 : 3.659 Too small

So x = 1.3 to 1dp

Equation for area: πx2 – 4 = x π x2 – x = 4?

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Work in 3s/4s (but you need to each individually show your working in your book)