Let's solve the given problem step by step:

a) Use the iterative formula to find an approximate solution (starting with $x_1 = 0$).

The iterative formula is: $x_{n+1} = \sqrt{\frac{3x_n + 12}{x_n + 2}}$ Starting with $x_1 = 0$, we'll compute successive iterations until the result stabilizes to 3 decimal places.

b) Calculate the difference between $x^3 + 2x^2 - 3x$ and 12 at the approximate solution found in part (a).

I will compute this now. math

Iterative formula function

def iterative_formula(x_n): return math.sqrt((3 * x_n + 12) / (x_n + 2))

Start with x1 = 0 and compute iterations

x_values = [0] # Initial guess tolerance = 0.001 # Stop when the change is less than this

while True: x_next = iterative_formula(x_values[-1]) x_values.append(x_next) if abs(x_next - x_values[-2]) < tolerance: break

Rounded result to 3 decimal places

approx_solution = round(x_values[-1], 3)

Part b: Calculate difference at the approximate solution

def calculate_difference(x): return abs((x3 + 2 * x2 - 3 * x) - 12)

difference = round(calculate_difference(approx_solution), 3)

approx_solution, difference### Results: a) The approximate solution to 3 decimal places is $x = 2.112$.
b) The difference between $x^3 + 2x^2 - 3x$ and 12 at $x = 2.112$ is $0.006$.

Would you like detailed steps for the iterative process or calculations?
Here are 5 related questions you might consider next:

1. What happens if the initial guess is different (e.g., $x_1 = 1$)?
2. How is the convergence rate of this iterative method determined?
3. Can we prove mathematically that the solution converges using this formula?
4. What happens if we use more decimal places in intermediate calculations?
5. How does this result compare to solving the equation using other methods, like Newton-Raphson?

Tip: Iterative methods are sensitive to starting values. Choosing a good initial guess can significantly speed up convergence!