https://file.aiquickdraw.com/mathbot/file/17303362627637g0zhm5m.jpg

To solve this using Newton's method, we need to apply the following iterative formula:

\[
x_{n+1} = x_n - \frac{f(x_n)}{f'(x_n)}
\]

where \( f(x) = 9x^2 + 3x - 1 \).

Let's break down the problem.

1. **Define \( f(x) \) and \( f'(x) \):**
   - \( f(x) = 9x^2 + 3x - 1 \)
   - \( f'(x) = 18x + 3 \)

2. **Set up initial guesses:**
   - Left solution: Start with \( x_0 = -1 \)
   - Right solution: Start with \( x_0 = 1 \)

3. **Compute \( x_1 \) and \( x_2 \) for each case:**

Let's go through these calculations.

---

### For the Left Solution \( (x_0 = -1) \)

1. **Compute \( x_1 \):**
   \[
   x_1 = x_0 - \frac{f(x_0)}{f'(x_0)}
   \]
   Substitute \( x_0 = -1 \):
   - \( f(-1) = 9(-1)^2 + 3(-1) - 1 = 9 - 3 - 1 = 5 \)
   - \( f'(-1) = 18(-1) + 3 = -18 + 3 = -15 \)

   So,
   \[
   x_1 = -1 - \frac{5}{-15} = -1 + \frac{1}{3} = -\frac{2}{3}
   \]

2. **Compute \( x_2 \):**
   Substitute \( x_1 = -\frac{2}{3} \):
   - \( f\left(-\frac{2}{3}\right) = 9\left(-\frac{2}{3}\right)^2 + 3\left(-\frac{2}{3}\right) - 1 = 9 \cdot \frac{4}{9} - 2 - 1 = 4 - 2 - 1 = 1 \)
   - \( f'\left(-\frac{2}{3}\right) = 18\left(-\frac{2}{3}\right) + 3 = -12 + 3 = -9 \)

   Then,
   \[
   x_2 = -\frac{2}{3} - \frac{1}{-9} = -\frac{2}{3} + \frac{1}{9} = -\frac{5}{9}
   \]

---

### For the Right Solution \( (x_0 = 1) \)

1. **Compute \( x_1 \):**
   Substitute \( x_0 = 1 \):
   - \( f(1) = 9(1)^2 + 3(1) - 1 = 9 + 3 - 1 = 11 \)
   - \( f'(1) = 18(1) + 3 = 18 + 3 = 21 \)

   So,
   \[
   x_1 = 1 - \frac{11}{21} = \frac{10}{21}
   \]

2. **Compute \( x_2 \):**
   Substitute \( x_1 = \frac{10}{21} \):
   - \( f\left(\frac{10}{21}\right) = 9\left(\frac{10}{21}\right)^2 + 3\left(\frac{10}{21}\right) - 1 = 9 \cdot \frac{100}{441} + \frac{30}{21} - 1 \approx 0.13 \)
   - \( f'\left(\frac{10}{21}\right) = 18\left(\frac{10}{21}\right) + 3 = \frac{180}{21} + 3 = \frac{243}{21} \approx 11.57 \)

   Then,
   \[
   x_2 \approx \frac{10}{21} - \frac{0.13}{11.57} \approx \frac{10}{21} - 0.0112 \approx 0.466
   \]

---

### Summary of Results
- For the left solution, \( x_2 \approx -\frac{5}{9} \).
- For the right solution, \( x_2 \approx 0.466 \).

Would you like further clarification or details on any steps? 

---

Here are some additional related questions to explore:

1. What are the general conditions for Newton's method to converge?
2. How would you interpret the derivative \( f'(x) \) in Newton's method?
3. How does the choice of the initial guess affect the convergence of Newton's method?
4. Can you find a situation where Newton's method does not converge?
5. What are other numerical methods for finding roots of equations?

**Tip:** Always check if the derivative \( f'(x) \) is close to zero, as it can lead to a very large next step, possibly diverging from the root.

Use Newton's method to estimate the solutions of the equation 9x^2 + 3x - 1 = 0. Start with x_0 = -1 for the left solution and x_0 = 1 for the right solution. Find x_2 in each case.

Learn how to estimate the solutions of the equation 9x^2 + 3x - 1 = 0 using Newton's method. Starting points include x_0 = -1 for the left solution and x_0 = 1 for the right solution, with detailed steps to find x_2. Suitable for students in Grades 11-12.

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