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# Trial and Improvement - PowerPoint PPT Presentation

Trial and Improvement. Objectives: C Grade Form and solve equations such as x 2 + x = 12 using trial and improvement. Prior knowledge: Rounding to decimal places Substitution into algebraic expressions The shape of quadratic / cubic graphs

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## PowerPoint Slideshow about ' Trial and Improvement' - kyrene

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Presentation Transcript

Objectives:

C Grade Form and solve equations such as x2 + x = 12

using trial and improvement

• Prior knowledge:

• Rounding to decimal places

• Substitution into algebraic expressions

• The shape of quadratic / cubic graphs

• Use of the bracket button on a calculator

Estimate the square root of the following numbers:

4.123105626

4.1

• 17

• 30

• 47

• 68

• 110

• 83

5.477225575

5.5

6.8556546

6.9

8.246211251

8.2

10.48808848

10.5

9.110433579

9.1

Find the positive solution to the equation

x2 - x = 60

y = x2 - x

If we consider this drawing a graph

we know the solution can be found by

drawing the line y = 60

and y = x2 – x, finding the value of x at

the point of intersection.

y = 60

because of the scale of the graph we

have to use we cannot find the value

of x to 1 d.p. but we can see it is

between 8 and 9.

Trial and improvement is where we try

values of x in the equation and try to

get as close to the given value for y

as possible

x2 – x = 60

Now we know that the solution is

between 8.2 and 8.3 we try 8.25

Try x = 8

x = 8 is too low

64 – 8= 56

Try x = 9

81 – 9= 72

x = 9 is too low

Try x = 8.5

72.25 – 8.5= 63.75

x = 8.5 is too high

Try x = 8.3

Now even the expanding

graph is not big enough

for the level

of accuracy required

68.89 – 8.3= 60.59

x = 8.3 is too high

Try x = 8.2

x = 8.2 is too low

67.42 – 8.2= 59.04

Trial and improvement is where we try

values of x in the equation and try to

get as close to the given value for y

as possible

Now we know that the solution is

between 8.2 and 8.3 we try 8.25

Try x = 8.25

x = 8.25 is too low

68.0625 – 8.25= 59.8125

We need the value of x to 1 d.p.

We know the solution is now between 8.25 and 8.3.

Try x = 8.3

Now even the expanding

graph is not big enough

for the level

of accuracy required

Any value between 8.25 and 8.3 would be rounded to 8.3

to 1 d.p

68.89 – 8.3= 60.59

x = 8.3 is too high

Try x = 8.2

Therefore to 1 d.p x = 8.3

x = 8.2 is too low

67.42 – 8.2= 63.75

Find the value of x to 1.d.p to solve this equation:

x3 + x = 12

We can now do this as a table:

27

3

3

30

Too High

8

2

2

10

Too Low

2.1

9.261

2.1

11.36

Too Low

12.85

10.65

2.2

2.2

Too High

9.938

2.15

12.09

2.15

Too High

9.8

2.14

2.14

11.94

Too Low

Because 2.15 is too high and any number less than 2.15 would be

rounded to 2.1 to 1 d.p.

Finding the answer when x = 2.14 proves that this is correct.

Now do these:

• Find the positive solutions to 1 decimal place

• 1. x2 + 2x = 63

• 2. x2 - 2x = 675

• 3. x3 + 2x = 520

• 4. x5 + x = 33 768

• 5. x2 - 7x = 368

• 6. x - x3 = -336

Worksheet

x = 7

x = 8.0

x = 27

x = 23

x = 8.0

x = 7.3