Show an equation has a solution in an interval, rearrange it into x=…, and repeat a formula to home in on the root. It is Higher only, aimed at grade 7. Free questions with answers are on this page.
Iteration is a recurrence xn+1=some formula in xn, starting from a given x0 or x1. Substitute, write the full calculator value, then round as asked for each term. To show a root lies in an interval, evaluate f(a) and f(b) and note the sign change. To show an iteration has converged 'to 2 d.p.', produce two consecutive terms that round to the same 2 d.p. value.
A diverging iteration grows or jumps around — say so if the terms are not settling. Do not round too early when feeding xn into the next step; use the unrounded display. Rearranging x3−3x+1=0 into x=33x−1 is how the formula appears; different rearrangements may or may not converge.
A sign change of f(a) and f(b) shows a root in (a,b). Rearrange to x=g(x), then xn+1=g(xn). Write every iterate to the required decimal places and look for a repeating digit.
Objectives
Show a sign change between two values and state the conclusion.
Reject an endpoint where f is already 0.
Run a square-root iteration and round the last value to 1 decimal place.
Key points
Show a root lies between two values by finding a sign change.
Rearrange an equation into the form x=g(x) and show the rearrangement is correct.
Use an iteration formula xn+1=g(xn) with a starting value, using the ANS key.
Write down each iterate to enough decimal places.
Worked examples
Example 1
Check the sign change for f(x) = x^2 + x - 5 by comparing f(1) and f(2). State the conclusion.
2 marks
f(1) = 1 + 1 - 5 = -3
f(2) = 4 + 2 - 5 = 1
The signs differ, so a root lies between 1 and 2
Answer: a root lies between 1 and 2
Common slips
Slip: f(1) = 1 - 5 = -4 The middle term is +x, so f(1) includes +1.
Slip: f(2) = 4 - 5 = -1 Add the 2 from the linear term.
Slip: The root is 1.5 You have shown a bracket, not the root itself.
Example 2
Use x = square root of (5 - x), starting at 2. Do 3 steps and give the result to 1 decimal place.
2 marks
First output √(5 - 2) = √3 ≈ 1.732
Next √(5 - 1.732) ≈ 1.808, then √(5 - 1.808) ≈ 1.787
1.787 to 1 decimal place is 1.8
Answer: 1.8
Common slips
Slip: √5 ≈ 2.236 Subtract the current x before you square-root.
Slip: Stop after one step The question asks for 3 steps.
Slip: 1.7 The following digit is 8, so round up.
Example 3
Check the sign of f(1) and f(2) for f(x) = x^2 + x - 8. State the conclusion about that interval.
3 marks
f(1) = 1 + 1 - 8 = -6
f(2) = 4 + 2 - 8 = -2
Both are negative, so there is no sign change between 1 and 2
Answer: no sign change between 1 and 2
Common slips
Slip: f(1) = 1 - 8 = -7 Include the +x term.
Slip: f(2) = 4 - 8 = -4 The linear term contributes +2.
Slip: A root lies between 1 and 2 Same signs do not trap a root in that gap.
Questions and answers
GCSE Maths iteration questions and answers. Try each one, then open the answer. Easy questions come first.
Question 1 Easy2 marks
Look at the sign of f(1) and f(2) for f(x) = x^2 + x - 4. What do you conclude?
Show answer
Answer: a root lies between 1 and 2
f(1) = -2 and f(2) = 2.
Opposite signs mean a root lies between 1 and 2.
M1 — f(1) = -2 and f(2) = 2.
A1 — Opposite signs mean a root lies between 1 and 2.
Question 2 Easy2 marks
For f(x) = x^2 + x - 3, compare the sign of f(1) with f(2) and state the conclusion.
Show answer
Answer: a root lies between 1 and 2
f(1) = -1 and f(2) = 3.
The sign changes, so a root lies between 1 and 2.
M1 — f(1) = -1 and f(2) = 3.
A1 — The sign changes, so a root lies between 1 and 2.
Question 3 Medium2 marks
Compare f(2) and f(3) for f(x) = x^2 + x - 8 and state the sign-change conclusion.
Show answer
Answer: a root lies between 2 and 3
f(2) = -2 and f(3) = 4.
The signs differ, so a root lies between 2 and 3.
M1 — f(2) = -2 and f(3) = 4.
A1 — The signs differ, so a root lies between 2 and 3.
Question 4 Medium2 marks
Sign-check f(2) and f(3) for f(x) = x^2 + x - 11. Write the conclusion.
Show answer
Answer: a root lies between 2 and 3
f(2) = -5 and f(3) = 1.
A root lies between 2 and 3.
M1 — f(2) = -5 and f(3) = 1.
A1 — A root lies between 2 and 3.
Question 5 Hard2 marks
f(x) = x^2 + x - 15. Check the sign of f(3) and f(4), and state where the root lies.
Show answer
Answer: a root lies between 3 and 4
f(3) = -3 and f(4) = 5.
A root lies between 3 and 4.
M1 — f(3) = -3 and f(4) = 5.
A1 — A root lies between 3 and 4.
Question 6 Hard2 marks
There is no sign change of f(x) = x^2 + x - 8 between f(1) and f(2). Say so in a sentence.
Show answer
Answer: no sign change between 1 and 2
f(1) = -6 and f(2) = -2.
Both are negative, so there is no sign change between 1 and 2.
M1 — f(1) = -6 and f(2) = -2.
A1 — Both are negative, so there is no sign change between 1 and 2.
Question 7 Exam-style5 marks
Use the iteration x = square root of (9 - x), starting at 2. Carry out 3 steps and give the result to 1 decimal place. Also say why a sign change of x^2 + x - 9 between 2 and 3 agrees with that.
Show answer
Answer: 2.5
√(9 - 2) = √7 ≈ 2.646.
Next ≈ 2.521, then ≈ 2.545.
2.545 to 1 decimal place is 2.5.
f(2) = -3 and f(3) = 3, so a root lies between 2 and 3.
The iteration is settling inside that interval.
M1 — First square root √7
M1 — Second step
M1 — Third step
A1 — 2.5
B1 — Sign change between 2 and 3 noted in the working
Common mistakes
Most lost marks on this topic are one slip.
Saying the root is halfway along a sign-change interval
The sign change only traps the root. Iteration or a formula gives the value
Treating f(1) = 0 as a sign change between 1 and 2
Zero at an endpoint means the root is that endpoint
Rounding each iteration before the next substitution
Round only the final output
Spot the slip
f(1) = 0 and f(2) = 4 for f(x) = x^2 + x - 2. A student says there is a sign change between 1 and 2. What is wrong?
f(1) is already a root, so the ends do not change sign
The values are too small
You need f(3) as well
Show the slip
f(1) is already a root, so the ends do not change sign One end is zero. A sign-change argument needs one negative value and one positive value.
Exam tips
A calculator is useful here. Show the calculation you typed, not only the final line.
If the question says “show that”, the result is given. The marks are for the steps that reach it.
The method is the same on AQA-style and Edexcel-style papers.
For 'show that … can be rearranged', start from the original equation and show every algebra line ending with the given form.
Write each iterate on its own line with its label (x1=…); every correct value can earn credit.
Printable worksheet
The same questions are in a PDF you can print. Answers start on a later page.