1. Approximate Solutions To Equations Using Iteration
Grade 9
Approximate Solutions To Equations Using Iteration
Find approximate solutions to equations using iteration, including using suffix notation in recursive formulae
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2. Lesson Plan Lesson Overview Progression of Learning
Objective(s)
Find approximate solutions to equations using iteration, including using suffix notation in recursive formulae
Grade 9
Prior Knowledge
Rearranging
General iterative processes (grade 7)
Duration
50 minutes
Resources
Print slides:
Equipment
Progression of Learning
What are the students learning?
How are the students learning? (Activities & Differentiation)
Rearranging an equation into an iterative form.
Give students slide 16. Using slides 4 and 5 explain how can rearrange in more than one way. 5
Solve – find approximate solutions using the iteration.
By substitution into the iterative formula for find the approximate solution by looking for 2 consecutive iterations to agree according to the question (e.g. 2 decimal places). Demonstrate using slide 6.
Students to practice. Give students slide 17 printed. 15
Find approximate solutions to equations using iteration in contextualised problems
Give students slide 18 printed. Allow students to work independently and then review the answer collectively using slide 9. 10
Find approximate solutions to equations using iteration in exam questions (from specimen papers)
Give students slides 19 and 20. This includes 5 exam questions related to objective. Students need to use notes from lesson to answer the questions. Ensure that all steps are shown. Relate to mark scheme to show how the marks are allocated. 20
Next Steps
Assessment
PLC/Reformed Specification/Target 7/Ratio, Proportion & Rates of Change/Approximate Solutions to Equations Using Iteration

3. Key Vocabulary
Iteration Recursive Solutions Substitute Rearrange Converge

4. Solving equations using iteration
Iteration is a method we can use to solve difficult equations such as x3 – x + 7=0.
Iterative formula is based on a recursive sequence. It
means using the previous term (n) to find the
next term (n+1) to get closer and closer to the
root of the equation.
An iterative rearrangement of x³-x+7 = 0 could be xn+1 = (xn3 + 7)

5. The Method – Part 1 b) x3 – x + 7 = 0 x3 – x + 7 = 0 x = x3 + 7
To use iteration we must rearrange an equation to the form x=g(x).
We can arrange it in more than one way.
Lets rearrange x3 – x + 7 = 0
Rearrange to isolate one of the x’s (could be the x or the x³)
x3 – x + 7 = 0
x = x3 + 7
xn+1 = (xn3 + 7)
b) x3 – x + 7 = 0
x3 = x - 7
x = 3√(x – 7)
xn+1 = 3√(xn – 7)
Now we have two iterative formulae

6. The Method – Part 2 xn+1 = 3√(xn – 7)
Step 1: Use the iterative formula with x0=-2.5
x1 = 3√(x0 – 7)
x1 = 3√(-2.5 – 7)
Step 2: Now use x1 to find x2,x3,x4,x5,x6 etc.
x1 =
x2 =
Step 3: Keep repeating the process under the result converges to the root.
x3 =
x4 =
x5 =
x6=
Step 4: The answer is to 4.d.p

7. Practice
Find the first five iterations of these iterative formulae starting with x1 = 2 a) xn+1 = 3xn + 6
b) xn+1 = 1-2xn
c) xn+1 = xn – 6 3
Starting with x0 = 3 show that the solution to the iterative formula
is x=2.99 correct to 2.d.p.
3. a) Show that the equation x² -5x -3=0 can be rearranged into x= √(5x+3).
b) Use the iterative formula xn+1 = √(5xn +3) starting with x0= 5 to find the solution correct to 1.d.p.

8. Practice - Solutions
Find the first five iterations of these iterative formulae starting with x1 = 2 a) xn+1 = 3xn + 6
b) xn+1 = 1-2xn
c) xn+1 = xn – 6 3
x2 = 12, x3= 42 , x4 = 132, x5 = 402, x6 = 1,212
b) x2 = -3, x3= 7 , x4 = -13, x5 = 27, x6 = -53
x2 = , x3= , x4 = ,
x5 = , x6 =
Starting with x0 = 3 show that the solution to the iterative formula
is x=2.99 correct to 2.d.p.
x1 = , x2= , x3 = , x4 =
As x3= x4 then answer is x=2.99 correct to 2.d.p.
3. a) Show that the equation x² -5x -3=0 can be rearranged into x= √(5x+3).
b) Use the iterative formula xn+1 = √(5xn +3) starting with x0= 5 to find the solution correct to 1.d.p.
b) x1 =
x2=
x3=
x4 =
x5 =
x6 = Answer 5.5 to 1.d.p
a) x2 – 5x -3 = 0
x2 = 5x + 3
x = √(5x +3)

9. Problem Solving and Reasoning
The volume of a cuboid is 170cm³. The dimensions of the cuboid are x, x and x+5. Use iteration to find the dimensions of the cuboid.
Step 1: Form an equation
Step 2: Make equation into iterative formula by rearranging for x.
x(x+5)x = 170
x³ = 170-5x²
x(x²+5x) = 170
x³ = x²
x³+5x² = 170
Step 3: Start with x1 = 4
x8=
x2 =
x9=
x3 =
x10=
Therefore x=4.3 to 1.dp
x4 =
x11=
x5 =
x12=
x6=
x13=
x7=
x14=

10. Reason and Explain
Explain why solving x³+x=3 is difficult? Will every iteration converge? Explain. Why is the graph of the equation useful in iteration? Does it matter what number you start with? Explore

11. Exam Questions – Specimen Papers

12. Exam Questions – Specimen Papers

13. Exam Questions – Specimen Papers

14. Exam Questions – Specimen Papers

15. Exam Questions – Specimen Papers

16. The Method
To use iteration we must rearrange an equation to the form x=g(x).
Lets rearrange x3 – x + 7 = 0
Student Sheet 1

17. Practice
Find the first five iterations of these iterative formulae starting with x1 = 2 a) xn+1 = 3xn + 6
b) xn+1 = 1-2xn
c) xn+1 = xn – 6 3
Starting with x0 = 3 show that the solution to the iterative formula
is x=2.99 correct to 2.d.p.
3. a) Show that the equation x² -5x -3=0 can be rearranged into x= √(5x+3).
b) Use the iterative formula xn+1 = √(5xn +3) starting with x0= 5 to find the solution correct to 1.d.p.
Student Sheet 2

18. Problem Solving and Reasoning
The volume of a cuboid is 170cm³. The dimensions of the cuboid are x, x and x+5. Use iteration to find the dimensions of the cuboid.
Step 1: Form an equation
Step 2: Make equation into iterative formula by rearranging for x.
Step 3: Start with x1 = 4
Student Sheet 3

19. Exam Questions – Specimen Papers - 1
Student Sheet 4

20. Exam Questions – Specimen Papers - 2
Student Sheet 5