Solving equations using iteration – Higher tier - Solving quadratic equations - Eduqas - GCSE Maths Revision - Eduqas

Solving equations using iteration – Higher tier

Approximate solutions to more complex equations can be found using a process called iteration. Iteration means repeatedly carrying out a process. To solve an equation using iteration, start with an intial value and substitute this into the equation to obtain a new value, then use the new value for the next substitution, and so on.

Example

Find the solution to the equation \(x^3 + 5x = 20\) using the initial value \(x_0=2\), giving the answer to 3 decimal places.

First, rearrange the equation to leave \(x\) on its own on one side of the equation.

One way to do this is:

\(x^3 + 5x = 20\)

\(x^3 = 20 - 5x\)

\(x = \sqrt[3]{20 - 5x}\)

To solve the equation, use the iterative formula \(x_{n+1} \sqrt[3]{20-5x_n}\)

We are given the initial value \(x_0=2\)

Substituting this into the iterative formula gives \(x_1 = \sqrt[3]{20-5 \times 2} = \sqrt[3]{10} = 2.154...\)

Substituting iteratively gives:

\(x_2 = \sqrt[3]{20-5 \times 2.154} = \sqrt[3]{9.227...} = 2.097\) (3dp)

\(x_3 = \sqrt[3]{20-5 \times 2.097} = \sqrt[3]{9.512...} = 2.118\) (3dp)

\(x_4 = \sqrt[3]{20-5 \times 2.118} = \sqrt[3]{9.405...} = 2.111\) (3dp)

\(x_5 = \sqrt[3]{20-5 \times 2.111} = \sqrt[3]{9.445...} = 2.114\) (3dp)

\(x_6 = \sqrt[3]{20-5 \times 2.114} = \sqrt[3]{9.430...} = 2.113\) (3dp)

\(x_7 = \sqrt[3]{20-5 \times 2.113} = \sqrt[3]{9.436...} = 2.113\) (3dp)

Since \(x_6\) and \(x_7\) give the same value to 3 decimal places, we can stop the iteration. The solution to the equation \(x^3 + 5x = 20\) is 2.113 to 3 decimal places.

Example

Show that the equation \(x^3+2x=7\) has a solution that lies between 1 and 2.

Use the iterative equation \(x_{n+1} = \sqrt[3]{7-2x_n}\) to find the solution to the equation to 3 decimal places.

Substituting \(x = 1\) into the left hand side of the equation gives \(1^3+2×1=3\)

Substituting \(x = 2\) into the left hand side of the equation gives \(2^3+2×2=12\)

The value on the right hand side of the equation, 7, lies between these two values, 3 and 12, so the solution to the equation must be between 1 and 2.

Using the initial value \(x_0=1\) and substituting this iteratively gives:

\(x_1 = \sqrt[3]{7-2\times1} = \sqrt[3]{5} = 1.710\) (3dp)

\(x_2 = \sqrt[3]{7-2\times1.710}... = \sqrt[3]{3.580}... = 1.530\) (3dp)

\(x_3 = \sqrt[3]{7-2\times1.530}... = \sqrt[3]{3.940}... = 1.579\) (3dp)

\(x_4 = \sqrt[3]{7-2\times1.579}... = \sqrt[3]{3.841}... = 1.566\) (3dp)

\(x_5 = \sqrt[3]{7-2\times1.566}... = \sqrt[3]{3.867}... = 1.570\) (3dp)

\(x_6 = \sqrt[3]{7-2\times1.570}... = \sqrt[3]{3.860}... = 1.569\) (3dp)

\(x_7 = \sqrt[3]{7-2\times1.569}... = \sqrt[3]{3.862}... = 1.569\) (3dp)

So the solution to the equation \(x^3+2x=7\) is 1.569 to 3 decimal places.

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