To solve this problem using the iterative process provided, we’ll follow the instructions closely to find an approximate solution for the equation $x^3 - 5x^2 - 12 = 0$ to two decimal places.

Iterative Process

1. Start with an initial value of $x = 5$.
2. Use the formula:
   $x_{\text{new}} = 5 + \frac{12}{x^2}$
3. Substitute $x = 5$ into this formula to get the next approximation. Then use the result as the new $x$-value in the next iteration and repeat until the values stabilize to two decimal places.

Step-by-Step Calculations

Let's go through a few iterations:

Iteration 1:

$x = 5$ $x_{\text{new}} = 5 + \frac{12}{5^2} = 5 + \frac{12}{25} = 5 + 0.48 = 5.48$

Iteration 2:

$x = 5.48$ $x_{\text{new}} = 5 + \frac{12}{5.48^2} \approx 5 + \frac{12}{30.0304} \approx 5 + 0.4 = 5.40$

Iteration 3:

$x = 5.40$ $x_{\text{new}} = 5 + \frac{12}{5.40^2} \approx 5 + \frac{12}{29.16} \approx 5 + 0.41 = 5.41$

Iteration 4:

$x = 5.41$ $x_{\text{new}} = 5 + \frac{12}{5.41^2} \approx 5 + \frac{12}{29.2681} \approx 5 + 0.41 = 5.41$

Conclusion

After a few iterations, we can see that the values converge to approximately 5.41. Therefore, the approximate solution to $x^3 - 5x^2 - 12 = 0$ to two decimal places is: $x \approx 5.41$

Would you like more details on any part of this process or have any questions?

Here are some related questions to explore further:

1. How can we verify that $x = 5.41$ is a solution to the equation $x^3 - 5x^2 - 12 = 0$?
2. What other iterative methods could be used to solve cubic equations?
3. How does the initial guess $x = 5$ affect the convergence of the iterative process?
4. Could this iterative formula fail to converge for some initial values? If so, why?
5. What are some practical applications of iterative methods in solving nonlinear equations?

Tip: When using iterative methods, monitoring the change in values between iterations helps to know when convergence is reached.