1. **State the problem:** Solve the cubic equation $$x^3 - 5x^2 - 12 = 0$$ using the iterative method defined by $$x_{n+1} = 5 + \frac{12}{x_n^2}$$ with initial guess $$x_1 = 5$$. 2. **Explain the iterative formula:** The iteration formula is derived by rearranging the original equation to isolate $$x$$ on one side: $$x^3 - 5x^2 - 12 = 0 \implies x^3 = 5x^2 + 12 \implies x = 5 + \frac{12}{x^2}$$ This formula will be used to generate successive approximations to the root. 3. **Perform iterations:** - Start with $$x_1 = 5$$. - Calculate $$x_2 = 5 + \frac{12}{5^2} = 5 + \frac{12}{25} = 5 + 0.48 = 5.48$$. - Calculate $$x_3 = 5 + \frac{12}{(5.48)^2} = 5 + \frac{12}{30.0304} \approx 5 + 0.3996 = 5.3996$$. - Calculate $$x_4 = 5 + \frac{12}{(5.3996)^2} = 5 + \frac{12}{29.157} \approx 5 + 0.4117 = 5.4117$$. - Calculate $$x_5 = 5 + \frac{12}{(5.4117)^2} = 5 + \frac{12}{29.298} \approx 5 + 0.4097 = 5.4097$$. 4. **Check convergence:** The values are approaching approximately $$5.41$$, so the root near this value is approximately $$x \approx 5.41$$. 5. **Verify root:** Substitute $$x=5.41$$ into the original equation: $$5.41^3 - 5 \times 5.41^2 - 12 \approx 158.3 - 146.4 - 12 = -0.1$$, close to zero, confirming the approximation. **Final answer:** The root of the equation near the initial guess is approximately $$\boxed{5.41}$$.