GCSE Maths · Higher · Chapter 24: Algebraic fractions and functions · Lesson 4

Iteration

Grades 7–9About 50 minutesCalculator allowed40 marks

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NameClassDate

Do Now

Grades 4–6

The first question is what today's lesson needs. The others bring back earlier topics.

A1
Needed today
Between which two whole numbers is the solution of $x^{3} = 20$ ?
[1 mark]
Answer
A2
From chapter 18
A class of width 5 has frequency 10. Find its frequency density.
[1 mark]
Answer
A3
From chapter 23
Where does $y = 4 \times 2^{x}$ cross the $y$ -axis?
[1 mark]
Answer
A4
From chapter 24
Make $x$ the subject of $k x + 9 = 5 x + y$ .
[1 mark]
Answer

Grades 7–8

Your teacher works through each example with you. Then try the one beside it.

Example 1

Use $x_{n + 1} = 3 x_{n} + 11$ with $x_{1} = 4$ to find $x_{2}$ , $x_{3}$ and $x_{4}$ , correct to 4 decimal places.

Hint Put each answer back into the formula. Keep the full calculator value: only round what you write down.

B1
Use $x_{n + 1} = 3 8 x_{n} - 3$ with $x_{1} = 3$ to find $x_{2}$ , $x_{3}$ and $x_{4}$ , correct to 4 decimal places.
[3 marks]
Answer $x_{2} = \dots\dots$ $x_{3} = \dots\dots$ $x_{4} = \dots\dots$

Example 2

Show that $x^{3} + 3 x - 20 = 0$ has a solution of $2.35$ , correct to 2 decimal places.

Hint Every number from

2.345

up to (not including)

2.355

rounds to

2.35

. A sign change between those two ends puts the solution inside.

B2
Show that $x^{3} - 5 x - 11 = 0$ has a solution of $2.95$ , correct to 2 decimal places.
[3 marks]

Grades 7–8

C1
Show that $x^{3} - 6 x - 3 = 0$ has a solution between $2$ and $3$ .
[2 marks]
C2
Show that $x^{3} - 3 x^{2} - 4 = 0$ can be rearranged to give $x = 3 + \frac{4}{x ^{2}}$ .
[2 marks]

C3
Use $x_{n + 1} = 3 20 - 2 x_{n}$ with $x_{1} = 2$ . Keep going until two values in a row are the same when rounded to 3 decimal places. Write down that solution, correct to 3 decimal places.
[3 marks]
Answer
C4
The values of $x_{n + 1} = \frac{2 x _{n} + 7}{3}$ settle on a solution of a quadratic equation. Write that equation in the form $a x^{2} + b x + c = 0$ , where $a$ , $b$ and $c$ are whole numbers.
[2 marks]
Answer
C5
Here are three rearrangements of $x^{3} - 5 x + 1 = 0$ . Which two are correct?
A: $x = 3 5 x - 1$ B: $x = \frac{x ^{3} - 1}{5}$
C: $x = 5 - \frac{1}{x}$
[2 marks]
Answer

Grades 7–9

D1
Sam solves $x^{2} - 5 x - 6 = 0$ with $x_{n + 1} = 5 x_{n} + 6$ and $x_{1} = 1$ . The values settle on $6$ . Sam says, "So $6$ is the only solution."
Explain why Sam is wrong.
[2 marks]

Grade 9 and beyond

No route is given. The hints are at the end of the sheet. Use one at a time.

Colour one circle on each line.

	Not yet	Nearly	Yes
I can rearrange an equation into an iterative formula and run it.
I can use a sign change to show where a solution lies.
I can prove a solution is correct to 2 decimal places.

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