Sign-Change Methods (WJEC Eduqas GCSE Maths): Revision Note
Exam code: C300
Sign-Change Methods
What is iteration?
Iteration means finding an approximation to the solution using a method that gets closer and closer (converges) to the real solution
It is often used for equations which don't have nice solutions
solutions themselves might be irrational numbers
How do I show that there is a solution between two values?
To find x0 (the initial / starting value), you are often asked to show that there is a solution between two values
For example, show that there is a solution to x3 + x = 7 between 1 and 2
Method 1:
Leave a constant term (e.g. the 7) on the right
substitute x = 1 and x = 2 into the left and show that this gives values below and above 7
13 + 1 = 2 and 23 + 2 = 10 which are below and above 7
A solution therefore lies between 1 and 2
Method 2:
Use "0" as your constant term on the right (by rearranging the equation into "... = 0")
then substitute in x = 1 and x = 2, showing this gives values below and above 0, i.e. negative and positive
this is called a change of sign between 1 and 2
x3 + x - 7 = 0
Substitute x = 1 into the left-hand side: 13 + 1 - 7 = -5 (negative)
Substitute x = 2 into the left-hand side: 23 + 2 - 7 = 3 (positive)
A solution lies between 1 and 2 as there is a change of sign
How do I find an approximate solution to an equation?
Trial and improvement
Write the equation in the form
Substitute values into the equation until you get two values
and
where:
Try values between
and
until you get two consecutive numbers at the correct degree of accuracy
e.g. if you need the answer to 1 decimal place then your two values should be 0.1 apart
Test the midpoint of these values to decide which one is closer to the solution
e.g. if
and
then test
If
then the solution is bigger than 3.25 so choose
If
then the solution is smaller than 3.25 so choose
Decimal search
Let’s say you know that a solution to an equation lies between 1 and 2 and you want to find it to 1 decimal place
The equation must be rearranged to have “= 0” on one side
Search every decimal, by substituting 1.0, 1.1, 1.2, 1.3, 1.4, … into the equation until you see a change of sign
Let’s say there was a change of sign between 1.6 and 1.7
This is not yet enough information to be able to round the solution to 1 decimal place
Substitute in 1.65, the value exactly in the middle of 1.6 and 1.7
If the sign change is between 1.6 and 1.65, the solution rounds down to 1.6 to 1 decimal place
If the sign change is between 1.65 and 1.7, the solution rounds up to 1.7 to 1 decimal place
Worked Example
A solution to the equation x3 + x = 7 lies between 1 and 2.
Find this solution correct to 1 decimal place.
Method 1: Decimal Search
Rearrange the equation to have = 0 on one side
x3 + x - 7 = 0
Substitute 1.0, 1.1, 1.2, 1.3, 1.4, … into the equation until a change of sign happens
x | x3 + x – 7 |
1.0 | -5 |
1.1 | -4.569 |
1.2 | -4.072 |
1.3 | -3.503 |
1.4 | -2.856 |
1.5 | -2.125 |
1.6 | -1.304 |
1.7 | -0.387 |
1.8 | 0.632 |
A change of sign is between 1.7 and 1.8 so test 1.75
When x = 1.75, x3 + x – 7 = 0.109375
See if the change of sign is between 1.7 and 1.75 or 1.75 and 1.8
The change of sign is between 1.7 and 1.75
The solution rounds to 1.7 to 1 decimal place
1.7 to 1 decimal place
Method 2: Interval Bisection
Rearrange the equation to have = 0 on one side
x3 + x - 7 = 0
1.5 bisects the interval 1 to 2
Substitute 1.0, 1.5 and 2.0 into the equation
x | x3 + x – 7 |
1.0 | -5 |
1.5 | -2.125 |
2.0 | 3 |
There is a change of sign between 1.5 and 2
Bisect this interval (1.75) and substitute this value into the equation
When x = 1.75, x3 + x – 7 = 0.109375
There is a change of sign between 1.5 and 1.75
Bisect this interval, (1.5 + 1.75) ÷ 2 = 1.625, and substitute this value into the equation
When x = 1.625, x3 + x – 7 = -1.084
There is a change of sign between 1.625 and 1.75
Bisect this interval, (1.625 + 1.75) ÷ 2 = 1.6875, and substitute this value into the equation
When x = 1.6875, x3 + x – 7 = -0.507...
There is a change of sign between 1.6875 and 1.75
This is enough to know that the solution rounds to 1.7 (it cannot be 1.75 itself as that gave 0.109375 from the equation)
1.7 to 1 decimal place
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