Question: An approximate solution to
$x^3 + 6x^2 + 9 = 0$ can be found
using this iterative formula:
$x_{n+1} = -6 - \frac{9}{x_n^2}$
Use this formula to work out an
approximate solution to the
equation, starting with $x_1 = -3.5$
Give your answer to 2 d.p.
1. **State the problem:** We want to find an approximate solution to the equation $$x^3 + 6x^2 + 9 = 0$$ using the iterative formula $$x_{n+1} = -6 - \frac{9}{x_n^2}$$ starting with $$x_1 = -3.5$$ and give the answer to 2 decimal places.
2. **Iterative formula:** The formula updates the value of $$x$$ by using the previous value $$x_n$$ to find $$x_{n+1}$$.
3. **Calculate $$x_2$$:**
$$x_2 = -6 - \frac{9}{(-3.5)^2} = -6 - \frac{9}{12.25} = -6 - 0.7347 = -6.7347$$
4. **Calculate $$x_3$$:**
$$x_3 = -6 - \frac{9}{(-6.7347)^2} = -6 - \frac{9}{45.35} = -6 - 0.1984 = -6.1984$$
5. **Calculate $$x_4$$:**
$$x_4 = -6 - \frac{9}{(-6.1984)^2} = -6 - \frac{9}{38.43} = -6 - 0.2343 = -6.2343$$
6. **Calculate $$x_5$$:**
$$x_5 = -6 - \frac{9}{(-6.2343)^2} = -6 - \frac{9}{38.87} = -6 - 0.2315 = -6.2315$$
7. The values are converging near $$-6.23$$.
**Final answer:** The approximate solution to 2 decimal places is $$\boxed{-6.23}$$.