https://file.aiquickdraw.com/mathbot/file/172841834974204c1pkeq.jpg

The image provides the following iterative formula to solve for \( x \):

\[
x_{n+1} = -4 - \frac{6}{x_n^2}
\]

The initial approximation for \( x \) is given as \( x_1 = -2.5 \). We are tasked with using this formula to find an approximate solution to the equation \( x^3 + 4x^2 + 6 = 0 \), and give the answer to 2 decimal places.

### Step-by-step Calculation:

1. **Iteration 1**:
   Using \( x_1 = -2.5 \):
   \[
   x_2 = -4 - \frac{6}{(-2.5)^2}
   \]
   \[
   x_2 = -4 - \frac{6}{6.25}
   \]
   \[
   x_2 = -4 - 0.96 = -4.96
   \]

2. **Iteration 2**:
   Using \( x_2 = -4.96 \):
   \[
   x_3 = -4 - \frac{6}{(-4.96)^2}
   \]
   \[
   x_3 = -4 - \frac{6}{24.60}
   \]
   \[
   x_3 = -4 - 0.24 = -4.24
   \]

3. **Iteration 3**:
   Using \( x_3 = -4.24 \):
   \[
   x_4 = -4 - \frac{6}{(-4.24)^2}
   \]
   \[
   x_4 = -4 - \frac{6}{17.98}
   \]
   \[
   x_4 = -4 - 0.33 = -4.33
   \]

Thus, after a few iterations, the value of \( x \) seems to stabilize around \( -4.33 \). Therefore, the approximate solution to the equation is:

\[
x \approx -4.33
\]

Would you like any further details on this process or more iterations? Let me know!

Here are 5 related questions:
1. How does the iterative method converge to a solution?
2. What happens if you start with a different initial guess for \( x_1 \)?
3. How can we verify the accuracy of this approximate solution?
4. What are some alternative methods for solving cubic equations like this one?
5. Why does this specific iterative method work for this equation?

**Tip:** Iterative methods often require a good initial guess to converge faster; choosing a poor initial guess can lead to slower convergence or divergence.

An approximate solution to x^3 + 4x^2 + 6 = 0 can be found using this iterative formula: x_{n+1} = -4 - 6/x_n^2. Use this formula to work out an approximate solution to the equation, starting with x_1 = -2.5. Give your answer to 2 d.p.

This page provides a step-by-step guide to solving the cubic equation x^3 + 4x^2 + 6 = 0 using the iterative formula x_{n+1} = -4 - 6/x_n^2. Starting with x_1 = -2.5, the solution is refined through iterations to an approximate value. Suitable for high school students learning numerical methods.

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