A20AlgebraHigherGrades 7–9

Iteration

There is no factorising trick for $x^{3} - 4x - 1 = 0$ . Instead, you can find an approximate solution by making a guess and repeatedly improving it. Each improvement is an $\textbf{iteration}$ , and a calculator does the repetitive work in seconds.

40 min AQA GCSE Maths

The big picture

The method has three stages. First show a solution exists between two values by finding a change of sign. Next rearrange the equation into the form $x = g(x)$ . Finally use $x_{n+1} = g(x_n)$ , feeding each answer back in, until the values settle.

AQA questions often give the iterative formula and ask you to show where it came from, use it for a few steps, and then justify the accuracy of the final answer.

What you'll learn

Show a root lies in an interval using a change of sign
Rearrange an equation into the form $x = g(x)$
Use an iterative formula to find approximate solutions
Justify an answer to a given degree of accuracy

Locating a root

If a continuous graph goes from negative to positive (or the other way) between two $x$ -values, it must cross the $x$ -axis between them.

For $f(x) = x^{3} - 4x - 1$ : $f(2) = 8 - 8 - 1 = -1$ and $f(3) = 27 - 12 - 1 = 14$ .

Example 1Show that

x^{3} - 4x - 1 = 0

has a solution between

x = 2

and

x = 3

f(2) = -1 < 0

f(3) = 14 > 0

AnswerThere is a change of sign, so there is a solution between 2 and 3.

Your turnSame method, new numbers — try it before moving on.

For $f(x) = x^3 - 5x + 1$ , work out $f(2)$ .

Your answer

Show worked solution

Solution

1 $f(2) = 8 - 10 + 1 = -1$ .
2(And $f(3) = 13$ , so there is a sign change between 2 and 3.)

Answer: $-1$

Tip — Write the conclusion in words. "Change of sign, so a root lies between…" is the mark.

Rearranging into x = g(x)

Isolate one $x$ term. From $x^{3} - 4x - 1 = 0$ : $x^{3} = 4x + 1$ , so $x = \sqrt[3]{4x + 1}$ .

Many rearrangements are possible; exam questions tell you which one to show.

Example 1Show that

x^{3} + 2x - 5 = 0

can be rearranged to

x = \sqrt[3]{5 - 2x}

x^{3} = 5 - 2x

2Cube root both sides.

Answer

x = \sqrt[3]{5 - 2x}

as required.

Using the iterative formula

Turn the rearrangement into $x_{n+1} = \sqrt[3]{4x_n + 1}$ and start from $x_0 = 2$ .

On a calculator: type 2, press =, then type the formula using ANS and press = repeatedly.

Example 1Use

x_{n+1} = \sqrt[3]{4x_n + 1}

with

x_0 = 2

to find

x_1

x_2

and

x_3

to 4 decimal places.

x_1 = \sqrt[3]{9} = 2.0801

x_2 = \sqrt[3]{9.3203} = 2.1045

x_3 = 2.1118

Answer

x_1 = 2.0801

x_2 = 2.1045

x_3 = 2.1118

Your turnSame method, new numbers — try it before moving on.

Using $x_{n+1} = \sqrt[3]{4x_n + 1}$ with $x_0 = 2$ , find $x_1$ to 4 decimal places.

Your answer

To 4 decimal places.

Show worked solution

Solution

1 $x_1 = \sqrt[3]{4 \times 2 + 1} = \sqrt[3]{9}$ .
2 $= 2.08008\ldots$ , so $2.0801$ .

Answer: $x_1 = 2.0801$

Why this works

The values climb towards about 2.114 and change less each time. Each iteration is a better estimate of where the graph of $y = x^{3} - 4x - 1$ crosses the $x$ -axis.

Justifying accuracy

To show a root is 2.115 to 3 decimal places, test the bounds 2.1145 and 2.1155.

$f(2.1145) = -0.0038$ and $f(2.1155) = 0.0056$ . The sign changes, so the root lies between them and rounds to 2.115.

Tip — Always test the bounds of the rounded answer, not the rounded answer itself.

Think like an examiner

Show the substituted values when checking for a sign change.
Write each iteration, not just the final one.
Use unrounded calculator values in the next iteration.

Remember these

f(a) < 0 < f(b)

Sign change: a root lies between a and b.

x_{n+1} = g(x_n)

Iterative formula.

x = \sqrt[3]{4x + 1}

A rearrangement of x cubed minus 4x minus 1 = 0.

Watch out for these

Rounding each iterate before using it again.

Use ANS to carry the full value forward.

Saying

f(2) = -1

is a solution.

A solution makes

f(x) = 0

; the sign change only locates it.

Testing 2.115 itself to justify 3 d.p.

Test 2.1145 and 2.1155.

Stating a sign change without a conclusion.

Say a root lies between the two values.

Stretch yourself

Show that $x^{3} + 2x - 5 = 0$ has a root between 1.3 and 1.4. Starting with $x_0 = 1$ , use $x_{n+1} = \sqrt[3]{5 - 2x_n}$ to find $x_4$ to 4 decimal places.

Hint — Evaluate $x^{3} + 2x - 5$ at 1.3 and 1.4 first.

Reveal solution

SolutionLet

f(x) = x^{3} + 2x - 5

f(1.3) = -0.203

and

f(1.4) = 0.544

: a change of sign, so a root lies between 1.3 and 1.4. Iterating:

x_1 = \sqrt[3]{3} = 1.4422

x_2 = 1.2837

x_3 = 1.3449

x_4 = 1.3220

. The values oscillate either side of the root, closing in on about 1.33.

\blacksquare

Questions students ask

Do iterations always get closer to the root?

No — some rearrangements move away. GCSE questions give formulas that work.

Which starting value should I use?

Use the one given; otherwise a value from the sign-change interval.

How many iterations are enough?

Keep going until successive values agree to the accuracy required.

Key takeaways

A sign change shows a root lies in an interval.
Rearrange $f(x) = 0$ into $x = g(x)$ .
Iterate with $x_{n+1} = g(x_n)$ using the ANS key.
Justify accuracy by testing the bounds.

Exam-style questions

Practise it the way it is examined

3 original questions · 8 marks, written to match the style and mark allocation of the real papers. Work on paper, then open the mark scheme and tick the marks you earned — method marks count even if the final answer slips.

Question 1Grade 72 marks

Show that the equation $x^3 - 6x - 4 = 0$ has a solution between $x = 2$ and $x = 3$ .[2]
Show mark scheme
- M1Evaluates both: $f(2) = -8$ and $f(3) = 5$
- A1States there is a change of sign, so a solution lies between 2 and 3

Question 2Grade 84 marks

(a)Using $x_{n+1} = \sqrt[3]{6x_n + 4}$ with $x_0 = 2.5$ , find $x_1$ , $x_2$ and $x_3$ to 4 decimal places.[2]
Show mark scheme (a)
- M1Correct first iteration: $x_1 = 2.6684$
- A1 $x_2 = 2.7149$ and $x_3 = 2.7274$
(b)Continue the iteration to find the solution correct to 2 decimal places.[2]
Show mark scheme (b)
- M1Further iterations showing convergence towards 2.732…
- A1 $2.73$
Answer: $2.73$

Question 3Grade 92 marks

Show that $x^3 - 2x^2 - 5 = 0$ can be rearranged to give $x = 2 + \dfrac{5}{x^2}$ .[2]
Show mark scheme
- M1Rearranges to $x^3 = 2x^2 + 5$
- A1Divides every term by $x^2$ to reach the given form

Unlimited practice

Skill drill

A new question every time, marked instantly, with a full worked solution. Aim for a streak of five, then move up a level.

0 / 0 first-time correct Streak 0Best at this level: 0

Generating a question…

Test yourself

Answer a few questions, then review anything you missed.

Next step

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