1. Trial and Improvement
13 April 2019

2. x – 1
Volume = 344 cm3
x + 4 x
………………………………….
WIDTH (x) = …………….cm

3. Finding the width of a rectangle
In the following rectangle the length is 1 cm more than the width. Its area is 82 cm2.
Let the width be x.
Choose a starting value for x
If x = 8
Area = 8 x 9 = 72 cm2
If x = 9
Area = 9 x 10 = 90 cm2
Choose a value to 1 d.p. between 8 and 9.
What decimal value of x is closest?
x = 8.6 cm
x + 1 x
Area = 82 cm2

4. A more usual exam type question follows:
Copy the table and you MUST show your trials.
Example x2 + x = 240
Trial x value
x2 + x
COMMENT 12
= 156
Too small 16
= ___

5. 1. x2 + x = 702
x value
x2 + x
COMMENT
2. x2 – x = 420
x value
x2 – x
COMMENT

6. Trial and Improvement
This method of solving an equation is called trial and improvement.
Its main features are:
Choose a whole number of x
Improve the choice by noting how far the answer is from what is needed.
If the answer is a decimal, find the value x to 1 decimal place.

7. 1. x2 + x = 66
x value
x2 + x
COMMENT
2. x2 – x = 133
x value
x2 – x
COMMENT