Sign-Change Methods (WJEC Eduqas GCSE Maths): Revision Note

Exam code: C300

Mark Curtis

Written by: Mark Curtis

Reviewed by: Dan Finlay

Updated on

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 straight f open parentheses x close parentheses equals 0

  • Substitute values into the equation until you get two values a and b where:

    • straight f open parentheses a close parentheses less than 0

    • straight f open parentheses b close parentheses greater than 0

  • Try values between a and b 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 straight f open parentheses 3.2 close parentheses less than 0 and straight f open parentheses 3.3 close parentheses greater than 0 then test x equals 3.25

      • If straight f open parentheses 3.25 close parentheses less than 0 then the solution is bigger than 3.25 so choose x equals 3.3

      • If straight f open parentheses 3.25 close parentheses greater than 0 then the solution is smaller than 3.25 so choose x equals 3.2

  • 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

Unlock more, it's free!

Join the 100,000+ Students that ❤️ Save My Exams

the (exam) results speak for themselves:

Mark Curtis

Author: Mark Curtis

Expertise: Maths Content Creator

Mark graduated twice from the University of Oxford: once in 2009 with a First in Mathematics, then again in 2013 with a PhD (DPhil) in Mathematics. He has had nine successful years as a secondary school teacher, specialising in A-Level Further Maths and running extension classes for Oxbridge Maths applicants. Alongside his teaching, he has written five internal textbooks, introduced new spiralling school curriculums and trained other Maths teachers through outreach programmes.

Dan Finlay

Reviewer: Dan Finlay

Expertise: Maths Subject Lead

Dan graduated from the University of Oxford with a First class degree in mathematics. As well as teaching maths for over 8 years, Dan has marked a range of exams for Edexcel, tutored students and taught A Level Accounting. Dan has a keen interest in statistics and probability and their real-life applications.