Solve x² - 2x - 8 = 0 using the fixed iteration method if the starting value of x is 5

The fixed-point iteration method is used to solve equations of the form \(x = g(x)\). To use this method for solving \(x^2 - 2x - 8 = 0\), we first need to rewrite the equation in the form \(x = g(x)\).

Starting from:
\[ x^2 - 2x - 8 = 0 \]
We can rearrange it to:
\[ x^2 - 2x = 8 \]
\[ x^2 = 2x + 8 \]
\[ x = \sqrt{2x + 8} \]

Thus, we define:
\[ g(x) = \sqrt{2x + 8} \]

Given the starting value \( x_0 = 5 \), we will use the fixed-point iteration method:
\[ x_{n+1} = g(x_n) = \sqrt{2x_n + 8} \]

We iterate this until the difference between successive values is less than a specified tolerance, say \( 10^{-5} \).

### Iteration Steps:

#### Iteration 1:
\[ x_1 = \sqrt{2x_0 + 8} = \sqrt{2 \cdot 5 + 8} = \sqrt{18} \approx 4.2426 \]

#### Iteration 2:
\[ x_2 = \sqrt{2x_1 + 8} = \sqrt{2 \cdot 4.2426 + 8} = \sqrt{16.4852} \approx 4.0582 \]

#### Iteration 3:
\[ x_3 = \sqrt{2x_2 + 8} = \sqrt{2 \cdot 4.0582 + 8} = \sqrt{16.1164} \approx 4.0139 \]

#### Iteration 4:
\[ x_4 = \sqrt{2x_3 + 8} = \sqrt{2 \cdot 4.0139 + 8} = \sqrt{16.0278} \approx 4.002 \]

#### Iteration 5:
\[ x_5 = \sqrt{2x_4 + 8} = \sqrt{2 \cdot 4.002 + 8} = \sqrt{16.004} \approx 4.0005 \]

#### Iteration 6:
\[ x_6 = \sqrt{2x_5 + 8} = \sqrt{2 \cdot 4.0005 + 8} = \sqrt{16.001} \approx 4.0001 \]

#### Iteration 7:
\[ x_7 = \sqrt{2x_6 + 8} = \sqrt{2 \cdot 4.0001 + 8} = \sqrt{16.0002} \approx 4.000 \]

Since the difference between \( x_7 \) and \( x_6 \) is less than \( 10^{-5} \), we can stop the iteration process.

Thus, the root of the equation \( x^2 - 2x - 8 = 0 \) using the fixed-point iteration method, starting from \( x_0 = 5 \), is approximately \( x \approx 4.0000 \).

Would you like more details or have any questions?

Here are some related questions for further exploration:
1. How do you choose a suitable function \( g(x) \) for fixed-point iteration?
2. What are the conditions for the convergence of the fixed-point iteration method?
3. How does the rate of convergence of fixed-point iteration compare with other iterative methods?
4. What are some common pitfalls to avoid when using fixed-point iteration?
5. Can fixed-point iteration be used for solving systems of nonlinear equations?
6. How does the choice of initial guess affect the convergence of fixed-point iteration?
7. What are the differences between fixed-point iteration and the Newton-Raphson method?
8. How can fixed-point iteration be modified to improve its convergence properties?

**Tip:** Always verify the convergence criteria and ensure the function \( g(x) \) is chosen such that the iterations remain bounded and converge to the desired root.

Learn how to solve the quadratic equation x² - 2x - 8 using the fixed-point iteration method with a step-by-step solution. This problem involves applying the fixed-point iteration technique to find the root of the quadratic equation, suitable for advanced high school students and above.

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