Trial and Improvement

Trial and ImprovementTrial and Improvement

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:

a. 17b. 30c. 47d. 68e. 110f. 83

4.15.56.98.2

10.59.1

Now check your answers on a calculator

4.1231056265.4772255756.85565468.246211251

10.488088489.110433579

y = x2 - x


Find the positive solution to the equation x2 - x = 60

give your answer to 1 decimal place

If we consider this drawing a graphwe know the solution can be found bydrawing the line y = 60and y = x2 – x, finding the value of x atthe point of intersection.

y = 60

because of the scale of the graph wehave to use we cannot find the valueof x to 1 d.p. but we can see it is between 8 and 9.


Trial and improvement is where we tryvalues of x in the equation and try toget as close to the given value for y

as possible

Try x = 8 x2 – x = 60

64 – 8 = 56 x = 8 is too low

Try x = 981 – 9 = 72 x = 9 is too low

Try x = 8.572.25 – 8.5 = 63.75 x = 8.5 is too high

Try x = 8.368.89 – 8.3 = 60.59 x = 8.3 is too high

Try x = 8.267.42 – 8.2 = 59.04 x = 8.2 is too low

Now even the expanding graph is not big enough

for the level of accuracy required

Now we know that the solution is between 8.2 and 8.3 we try 8.25


Trial and improvement is where we tryvalues of x in the equation and try toget as close to the given value for y

as possible

Try x = 8.2568.0625 – 8.25 = 59.8125 x = 8.25 is too low

Try x = 8.368.89 – 8.3 = 60.59 x = 8.3 is too high

Try x = 8.267.42 – 8.2 = 63.75 x = 8.2 is too low

Now even the expanding graph is not big enough

for the level of accuracy required

Now we know that the solution is between 8.2 and 8.3 we try 8.25

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

We know the solution is now between 8.25 and 8.3.

Any value between 8.25 and 8.3 would be rounded to 8.3to 1 d.p

Therefore to 1 d.p x = 8.3


Find the value of x to 1.d.p to solve this equation:x3 + x = 12

We can now do this as a table:

Trial Value of x

x3 x x3-xComment

3 27 3 30 Too High

2 8 2 10 Too Low

2.1 9.261 2.1 11.36 Too Low

2.2 10.65 2.2 12.85 Too High

2.15 9.938 2.15 12.09 Too High

2.14 9.8 2.14 11.94 Too Low

Because 2.15 is too high and any number less than 2.15 would berounded to 2.1 to 1 d.p.

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

Trial and ImprovementTrial and ImprovementNow do these:

Find the positive solutions to 1 decimal place1. x2 + 2x = 63

2. x2 - 2x = 675

3. x3 + 2x = 520

4. x5 + x = 33 768

5. x2 - 7x = 368

6. x - x3 = -336


x2 + 2x = 63 x5 + x = 33 768Trial Value of x

x2 2x x2 + 2x Comment

Trial Value of x

x5 x x5 + x Comment

x2 - 2x = 675 x2 - 7x = 368Trial Value of x

x2 2x x2 - 2x Comment

Trial Value of x


x3 + 2x = 520 x - x3 = -336Trial Value of x


Trial Value of x

x x3 x - x3 Comment

Worksheet


x2 + 2x = 63 63 x5 + x = 33 768 33768Trial Value of x


Trial Value of x

x5 x x5 + x Comment

5 25 10 35 Too Low 8 32768 16 32784 Too Low6 36 12 48 Too Low 9 59049 18 59067 Too High7 49 14 63 8.1 34868 16.2 34884 Too High 8.05 33805 16.1 33821 Too High 8.02 33180 16.04 33196 Too Low 8.04 33595 16.08 33612 Too Low

x2 - 2x = 675 675 x2 - 7x = 368 368Trial Value of x


Trial Value of x


20 400 40 360 Too Low 25 625 175 450 Too High30 900 60 840 Too High 22 484 154 330 Too Low28 784 56 728 Too High 23 529 161 368 26 676 52 624 Too Low 27 729 54 675

x3 + 2x = 520 520 x - x3 = -336 -336Trial Value of x


Trial Value of x

x x3 x - x3 Comment

8 512 16 528 Too High 7 49 343 -294 Too High7 343 14 357 Too Low 8 64 512 -448 Too Low

7.5 421.9 15 436.9 Too Low 7.5 56.25 421.9 -366 Too Low7.9 493 15.8 508.8 Too Low 7.4 54.76 405.2 -350 Too Low7.95 502.5 15.9 518.4 Too Low 7.3 53.29 389 -336 7.96 504.4 15.92 520.3 Too High

x = 7

x = 27

x = 8.0

x = 8.0

x = 23

x = 7.3

Trial and Improvement

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