1. Trial 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

2. Trial and Improvement
Estimate the square root of the following numbers: 17 30 47 68
110 83
4.1
5.5
6.9
8.2
10.5
9.1
Now check your answers on a calculator

3. Find the positive solution to the equation x2 - x = 60
y = x2 - x
Trial and Improvement
Find the positive solution to the equation
x2 - x = 60
give your answer to 1 decimal place
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.

4. 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
Try x = 8
Now we know that the solution is
between 8.2 and 8.3 we try 8.25
64 – 8 = 56
x = 8 is too low
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
67.42 – 8.2 = 59.04
x = 8.2 is too low

5. 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
– 8.25 =
x = 8.25 is too low
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
67.42 – 8.2 = 63.75
x = 8.2 is too low

6. Find the value of x to 1.d.p to solve this equation: x3 + x = 12
Trial and Improvement
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-x
Comment 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 be
rounded to 2.1 to 1 d.p.
Finding the answer when x = 2.14 proves that this is correct.

7. Trial and Improvement
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 =
5. x2 - 7x = 368
6. x - x3 = -336

8. Trial and Improvement Worksheet x2 + 2x = 63 x5 + x = 33 768
Trial Value of x x2 2x
x2 + 2x
Comment x5 x
x5 + x
x2 - 2x = 675
x2 - 7x = 368
x2 - 2x 7x
x2 - 7x
x3 + 2x = 520
x - x3 = -336 x3
x3 + 2x
x - x3

9. Trial and Improvement x = 7 x = 8.0 x = 27 x = 23 x = 8.0 x = 7.363
x5 + x =
33768
Trial Value of x x2 2x
x2 + 2x
Comment x5 x
x5 + x 5 25 10 35
Too Low 8
32768 16
32784 6 36 12 48 9
59049 18
59067
Too High 7 49 14
8.1
34868
16.2
34884
8.05
33805
16.1
33821
8.02
33180
16.04
33196
8.04
33595
16.08
33612
x2 - 2x = 675
675
x2 - 7x = 368
368
x2 - 2x 7x
x2 - 7x 20
400 40
360
625
175
450 30
900 60
840 22
484
154
330 28
784 56
728 23
529
161 26
676 52
624 27
729 54
x3 + 2x = 520
520
x - x3 = -336
-336 x3
x3 + 2x
x - x3
512
528
343
-294
357 64
-448
7.5
421.9 15
436.9
56.25
-366
7.9
493
15.8
508.8
7.4
54.76
405.2
-350
7.95
502.5
15.9
518.4
7.3
53.29
389
7.96
504.4
15.92
520.3
x = 8.0
x = 27
x = 23
x = 8.0
x = 7.3