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Maths · Equations and graphs

Using iteration to solve equations

Show that a root lies in an interval using a change of sign, use decimal search, and use iteration formulae \(x_{n+1} = f(x_n)\) to find roots to a given accuracy.

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Teacher resources

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Student handouts

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Warm-up

Answer each one, then check.

  1. 1

    Work out \(1^3 - 1 - 1\).

    Show answerHide answer

    \(-1\)

  2. 2

    Work out \(2^3 - 2 - 1\).

    Show answerHide answer

    \(5\)

  3. 3

    What is \(\sqrt[3]{8}\)?

    Show answerHide answer

    \(2\)

  4. 4

    What does ANS do on a calculator?

    Show answerHide answer

    Uses the previous answer

  5. 5

    What does "to 2 decimal places" mean?

    Show answerHide answer

    Rounded to two digits after the point

Learning Objectives

  1. 1Show that a root lies between two values.
  2. 2Use decimal search to narrow down a root.
  3. 3Use an iteration formula \(x_{n+1} = f(x_n)\).
  4. 4Rearrange an equation into an iteration formula.

CHANGE OF SIGN

If \(f(a)\) and \(f(b)\) have opposite signs, the graph crosses the x-axis between \(a\) and \(b\), so there is a root in that interval.

This works when \(f\) is continuous, with no breaks in the curve.

Change of Sign

Show that \(x^3 - x - 1 = 0\) has a root between 1 and 2.

Show the solutionHide the solution
  1. 1 Write the function \(f(x) = x^3 - x - 1\)
  2. 2 \(f(1)\) \(1 - 1 - 1 = -1\)
  3. 3 \(f(2)\) \(8 - 2 - 1 = 5\)
  4. 4 Sign change Negative to positive

Answer\(f(1) < 0\) and \(f(2) > 0\), so there is a root between 1 and 2.

Decimal Search

Find the root of \(x^3 - x - 1 = 0\) to 2 decimal places.

Show the solutionHide the solution
  1. 1 \(f(1.3)\) \(-0.103\) (negative)
  2. 2 \(f(1.4)\) \(0.344\) (positive)
  3. 3 \(f(1.32)\) \(-0.020\) (negative)
  4. 4 \(f(1.33)\) \(0.023\) (positive)
  5. 5 Midpoint \(f(1.325)\) \(0.001\) (positive), so the root is below 1.325

AnswerThe root is 1.32 to 2 decimal places.

Iteration

\(x_{n+1} = \sqrt[3]{x_n + 1}\) with \(x_0 = 1\). Work out \(x_1\), \(x_2\) and \(x_3\).

Show the solutionHide the solution
  1. 1 \(x_1\) \(\sqrt[3]{2} = 1.2599\)
  2. 2 \(x_2\) \(\sqrt[3]{1.2599 + 1} = 1.3123\)
  3. 3 \(x_3\) \(\sqrt[3]{1.3123 + 1} = 1.3224\)

Answer\(x_1 = 1.2599\), \(x_2 = 1.3123\), \(x_3 = 1.3224\); the values are approaching the root 1.3247.

Rearranging to an Iteration Formula

Show that \(x^3 - x - 1 = 0\) can be rearranged to \(x = \sqrt[3]{x + 1}\).

Show the solutionHide the solution
  1. 1 Add \(x + 1\) \(x^3 = x + 1\)
  2. 2 Cube root \(x = \sqrt[3]{x + 1}\)

Answer\(x^3 - x - 1 = 0\) gives \(x^3 = x + 1\), so \(x = \sqrt[3]{x + 1}\).

Calculator Tips

Use your calculator efficiently.

  • Type the starting value

    Press =, then type the formula using ANS.

  • Keep pressing =

    Each press gives the next term.

  • Do not round in between

    Use the full value from ANS.

  • Compare

    When two terms agree to the accuracy needed, stop.

Close In on a Root

(a) Show that \(x^3 + 2x - 7 = 0\) has a root between 1 and 2. (b) Use decimal search to find the root to 1 decimal place.

1. Test the midpoint of the last interval.

2. Decide which way to round.

A good answer shows: (a) \(f(1) = -4\), \(f(2) = 5\). (b) \(f(1.5) = -0.625\), \(f(1.6) = 0.296\), \(f(1.55) = -0.176\); the root is between 1.55 and 1.6, so it is 1.6 to 1 d.p.

Can I...?

  1. 1Evaluate \(f(a)\) and \(f(b)\).
  2. 2Explain a change of sign.
  3. 3Use decimal search.
  4. 4Use a starting value.
  5. 5Use ANS on a calculator.
  6. 6Use an iteration formula.
  7. 7Rearrange to an iteration formula.
  8. 8Stop at the required accuracy.

Summary & Exam Focus

  • Change of sign shows a root in an interval.
  • Decimal search narrows the interval.
  • Iteration: feed each answer back into the formula.
  • Stop when values agree to the required accuracy.

Exam focus

Show that \(x^3 - x - 1 = 0\) has a root between 1 and 2. (2 marks) (2 marks)

Show both values of \(f\) and state that the sign changes.

Key terms

The vocabulary this lesson expects you to use. Each one is linked from the first place it appears above.

Iteration
Repeating a calculation using the previous answer.
Iteration formula
A rule \(x_{n+1} = f(x_n)\).
Starting value
The first value, \(x_0\).
Change of sign
\(f(a)\) and \(f(b)\) have opposite signs.
Root
A value of \(x\) where \(f(x) = 0\).
Convergence
The values getting closer to a limit.

Questions and answers

12 questions set on this lesson, with the mark schemes and model answers open.

1. Exam question Show that 2 marks Easier

Show that the equation \(x^3 - x - 1 = 0\) has a root between 1 and 2.

Mark scheme — 2 marks available

  • \(f(1) = -1\) and \(f(2) = 5\) — M1
  • Change of sign conclusion — C1

Model answer

\(f(1) = -1\) and \(f(2) = 5\). There is a change of sign, so a root lies between 1 and 2.

2. Exam question Work out 3 marks Easier

\(x_{n+1} = \sqrt[3]{x_n + 1}\) and \(x_0 = 1\). Work out the values of \(x_1\), \(x_2\) and \(x_3\).

Mark scheme — 3 marks available

  • \(x_1 = 1.26\) — B1
  • \(x_2 = 1.31\) — B1
  • \(x_3 = 1.32\) — B1

Model answer

\(x_1 = 1.2599\), \(x_2 = 1.3123\), \(x_3 = 1.3224\)

3. Exam question Show that 2 marks Easier

Show that \(x^3 - x - 1 = 0\) can be rearranged to give \(x = \sqrt[3]{x + 1}\).

Mark scheme — 2 marks available

  • \(x^3 = x + 1\) — M1
  • Cube root both sides — A1

Model answer

\(x^3 = x + 1\), so \(x = \sqrt[3]{x + 1}\).

4. Exam question Find 4 marks Easier

The graph of \(y = x^3 - x - 1\) is shown. It crosses the x-axis between \(x = 1\) and \(x = 2\). Use a trial and improvement method to find this root correct to 1 decimal place. You must show all your working.

The graph of y equals x cubed minus x minus 1 between 0 and 2, crossing the x-axis near 1.3.

Mark scheme — 4 marks available

  • Tests values in the interval — M1
  • \(f(1.3) < 0\) and \(f(1.4) > 0\) — A1
  • Tests \(1.35\) — M1
  • 1.3 — A1

Model answer

\(f(1.3) = -0.103\); \(f(1.4) = 0.344\); the root is between 1.3 and 1.4. \(f(1.35) = 0.1104\) (positive), so the root is below 1.35 and the answer is 1.3.

5. Exam question Find 4 marks Easier

(a) Show that \(x^3 + 2x - 7 = 0\) has a root between 1 and 2. (b) Find this root correct to 1 decimal place.

Mark scheme — 4 marks available

  • \(f(1) = -4\) and \(f(2) = 5\) — B1
  • Tests 1.5 or 1.6 — M1
  • Tests 1.55 — M1
  • 1.6 — A1

Model answer

(a) \(f(1) = -4\), \(f(2) = 5\): change of sign. (b) \(f(1.5) = -0.625\), \(f(1.6) = 0.296\), \(f(1.55) = -0.176\), so the root is between 1.55 and 1.6: 1.6.

6. Exam question Explain 2 marks Easier

Using \(x_{n+1} = \sqrt[3]{x_n + 1}\) with \(x_0 = 1\), the values are \(x_5 = 1.32463\) and \(x_6 = 1.32470\). Write down the root of \(x^3 - x - 1 = 0\) to 3 decimal places and explain how you know.

Mark scheme — 2 marks available

  • 1.325 — B1
  • Successive values agree to 3 d.p. — C1

Model answer

1.325, because both \(x_5\) and \(x_6\) round to 1.325 to 3 decimal places.

7. Multiple choice 1 mark Easier

A change of sign of \(f(x)\) between 1 and 2 tells you...

  1. A The graph is a straight line
  2. B \(f(1.5) = 0\)
  3. C There is a root between 1 and 2 Correct
  4. D There is no root

Why: There is a root between 1 and 2.

8. Multiple choice 1 mark Core

\(f(x) = x^3 - x - 1\). \(f(1)\) is...

  1. A \(-1\) Correct
  2. B \(0\)
  3. C \(1\)
  4. D \(-3\)

Why: \(1 - 1 - 1 = -1\).

9. Multiple choice 1 mark Core

In an iteration \(x_{n+1} = f(x_n)\), \(x_0\) is called...

  1. A The root
  2. B The answer
  3. C The formula
  4. D The starting value Correct

Why: The starting value.

10. Multiple choice 1 mark Core

To use ANS on a calculator you...

  1. A Reset the calculator
  2. B Use the previous answer in the formula Correct
  3. C Delete the last answer
  4. D Type the root

Why: Type the formula with ANS in place of \(x_n\), then keep pressing equals.

11. Multiple choice 1 mark Core

You can stop iterating when...

  1. A The first value is a whole number
  2. B The value is negative
  3. C Successive values agree to the required accuracy Correct
  4. D After exactly 3 steps

Why: Successive values agree to the required accuracy.

12. Multiple choice 1 mark Stretch

Rearranging \(x^2 = 5x - 3\) into an iteration formula could give...

  1. A \(x_{n+1} = \sqrt{5x_n - 3}\) Correct
  2. B \(x_{n+1} = 5x_n - 3\)
  3. C \(x_{n+1} = x_n^2\)
  4. D \(x_{n+1} = 3 - 5x_n\)

Why: Take the square root: \(x = \sqrt{5x - 3}\).