To solve the problem using the iterative process described for the equation:

$x^3 - 5x^2 - 9 = 0$

The iteration formula provided is:

$x = 5 + \frac{9}{x^2}.$

We are starting with $x = 5$, and the process involves repeating the steps until the value of $x$ stabilizes (rounded to 2 decimal places). Let me calculate the approximate solution iteratively for you:

Step 1: Start with $x_0 = 5$.

Step 2: Use the iteration formula to find the next value:

$x_{n+1} = 5 + \frac{9}{x_n^2}.$

Iterations:

1. Iteration 1 (Starting with $x_0 = 5$): $x_1 = 5 + \frac{9}{5^2} = 5 + \frac{9}{25} = 5 + 0.36 = 5.36$

2. Iteration 2 (Using $x_1 = 5.36$): $x_2 = 5 + \frac{9}{(5.36)^2}.$ First, calculate $5.36^2$: $5.36^2 = 28.7296.$ Now: $x_2 = 5 + \frac{9}{28.7296} = 5 + 0.3132 = 5.3132 \quad (\text{rounded to 4 d.p.}).$

3. Iteration 3 (Using $x_2 = 5.3132$): $x_3 = 5 + \frac{9}{(5.3132)^2}.$ First, calculate $5.3132^2$: $5.3132^2 = 28.2339.$ Now: $x_3 = 5 + \frac{9}{28.2339} = 5 + 0.3189 = 5.3189.$

4. Iteration 4 (Using $x_3 = 5.3189$): $x_4 = 5 + \frac{9}{(5.3189)^2}.$ First, calculate $5.3189^2$: $5.3189^2 = 28.2866.$ Now: $x_4 = 5 + \frac{9}{28.2866} = 5 + 0.3182 = 5.3182.$

Rounding to 2 Decimal Places:

At this point:

• $x_3 = 5.32$
• $x_4 = 5.32$.

The values stabilize at $x = 5.32$ (to 2 d.p.).

Final Answer:

The approximate solution to $x^3 - 5x^2 - 9 = 0$ is:

$\boxed{5.32}$

Would you like a step-by-step explanation for any part of this process or further clarification?

Related Questions:

1. How can this iterative process be generalized for other equations?
2. What happens if we start with a different initial guess?
3. Can the iteration formula fail to converge? Under what conditions?
4. How is this iteration process related to fixed-point iteration?
5. How can we graphically interpret the convergence of the solution?

Tip:

The accuracy of the iterative process depends heavily on the starting value and the behavior of the function. Always check the convergence of the iterations.