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Approximating solutions to equations
I can use iteration to find approximate solutions to equations.
Lesson 4 of 6
New
New
- Year 11
Approximating solutions to equations
I can use iteration to find approximate solutions to equations.
Lesson details
Key learning points
- You can rearrange an equation in order to carry out iteration
- This process finds an approximate solution to the equation
- This process is carried out until the outputs are the same to a given degree of accuracy
- Efficient use of the calculator makes this process much faster
Keywords
Iteration - Iteration is the repeated application of a function or process in which the output of each iteration is used as the input for the next iteration.
Common misconception
To find a solution to a given degree of accuracy an iteration can just be rounded.
Several successive iterations need to round to the same value to a given degree of accuracy. If in doubt efficient calculator use means many iterations can be generated easily until confident that a digit is fixed.
Pupils can check solutions found by iteration using the equation feature on their calculator. Most of the equations in this lesson are cubic so pupils will be unable to solve in other ways.
Teacher tip
Licence
This content is © Oak National Academy Limited (2025), licensed on Open Government Licence version 3.0 except where otherwise stated. See Oak's terms & conditions (Collection 2).
Lesson video
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Prior knowledge starter quiz
6 Questions
Q1.What is 0.0647399 rounded to 2 significant figures?
What is 0.0647399 rounded to 2 significant figures?
0.1
0.064
Correct answer: 0.065
0.065
0.0647
6.47
Q2.What is the value of the expression $$x^3 +3x-12$$ when $$x=2$$?
What is the value of the expression $$x^3 +3x-12$$ when $$x=2$$?
Correct Answer: 2
2
Q3.For which values of $$x$$ are we able to evalate the expression $$\sqrt{6-2x}$$?
For which values of $$x$$ are we able to evalate the expression $$\sqrt{6-2x}$$?
Correct answer: -1
-1
Correct answer: 0
0
Correct answer: 2
2
Correct answer: 3
3
4
Q4.Use the iterative formula $$P_{t+1}=2.5(P_t-50)$$ with $$P_0=200$$ to find the value of $$P_4$$. Write down your full calculator display.
Use the iterative formula $$P_{t+1}=2.5(P_t-50)$$ with $$P_0=200$$ to find the value of $$P_4$$. Write down your full calculator display.
Correct Answer: 4640.625
4640.625
Q5.Using the iterative formula $$x_{n+1}=10-4(x_n)$$ with $$x_0=1$$ which of these is the value of $$x_8$$?
Using the iterative formula $$x_{n+1}=10-4(x_n)$$ with $$x_0=1$$ which of these is the value of $$x_8$$?
$$-22$$
$$16\;386$$
Correct answer: $$-65\;534$$
$$-65\;534$$
$$262\;146$$
Q6.Using the iterative formula $$x_{n+1}=\sqrt{10-2(x_n)}$$ and $$x_0=3$$ match up the iterations with their values.
Using the iterative formula $$x_{n+1}=\sqrt{10-2(x_n)}$$ and $$x_0=3$$ match up the iterations with their values.
Correct Answer:$$x_1$$,2
$$x_1$$ -Â 2
Correct Answer:$$x_2$$,2.449489...
$$x_2$$ -Â 2.449489...
Correct Answer:$$x_3$$,2.258543...
$$x_3$$ -Â 2.258543...
Correct Answer:$$x_4$$,2.341561...
$$x_4$$ -Â 2.341561...
Correct Answer:$$x_5$$,2.305835 ...
$$x_5$$ -Â 2.305835 ...
Assessment exit quiz
6 Questions
Q1.Which value does the iterative formula $$x_{n+1}=\frac{5}{x_n}-3$$ converge on to 1 significant figure when $$x_0=2$$ ?
Which value does the iterative formula $$x_{n+1}=\frac{5}{x_n}-3$$ converge on to 1 significant figure when $$x_0=2$$ ?
-10
Correct answer: -4
-4
-3
-0.5
Q2.Which value does the iterative formula $$x_{n+1}=\frac{5}{x_n}-3$$ converge on to 2 significant figures when $$x_0=2$$ ?
Which value does the iterative formula $$x_{n+1}=\frac{5}{x_n}-3$$ converge on to 2 significant figures when $$x_0=2$$ ?
-4.5
-4.4
Correct answer: -4.2
-4.2
-4.1
Q3.How can we tell if an iterative formula is converging for a given value of $$x_0$$?
How can we tell if an iterative formula is converging for a given value of $$x_0$$?
The iterations alternate between positive and negative.
The iterations are increasing each time.
The iterations are decreasing each time.
Correct answer: The difference between iterations is decreasing.
The difference between iterations is decreasing.
Q4.Which of these could be an iterative formula to find a solution to $$x^3-3x-5=0$$ ?
Which of these could be an iterative formula to find a solution to $$x^3-3x-5=0$$ ?
$$x_{n+1}=\frac{(x_n)^3}{3} - 5$$
$$x_{n+1}=\frac{(x_n)^3+5}{3}$$
$$x_{n+1}=\sqrt[3]{-3(x_n)-5}$$
Correct answer: $$x_{n+1}=\sqrt[3]{3(x_n)+5}$$
$$x_{n+1}=\sqrt[3]{3(x_n)+5}$$
Q5.To find a solution to the equation $$x^3+6x-10=0$$ the iterative formula $$x_{n+1}=\frac{10-(x_n)^3}{6}$$ can be used. Using $$x_0=1$$ find the value of $$x_3$$
To find a solution to the equation $$x^3+6x-10=0$$ the iterative formula $$x_{n+1}=\frac{10-(x_n)^3}{6}$$ can be used. Using $$x_0=1$$ find the value of $$x_3$$
-2.8333...
1.1041...
1.1666...
Correct answer: 1.4423...
1.4423...
Q6.Using the iterative formula $$x_{n+1}=\frac{10-(x_n)^3}{6}$$ and $$x_0=1$$ there is a solution to $$x^3+6x-10=0$$ correct to 2 significant figures when $$x=$$
Using the iterative formula $$x_{n+1}=\frac{10-(x_n)^3}{6}$$ and $$x_0=1$$ there is a solution to $$x^3+6x-10=0$$ correct to 2 significant figures when $$x=$$
Correct Answer: 1.3
1.3
