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To solve this problem, let's use **Newton's Method** to approximate the roots of the equation:

\[
9x^2 + 4x - 1 = 0
\]

Newton's Method formula is:
\[
x_{n+1} = x_n - \frac{f(x_n)}{f'(x_n)}
\]
where \( f(x) = 9x^2 + 4x - 1 \), and we need to compute \( f'(x) \), the derivative of \( f(x) \).

### Step 1: Find \( f(x) \) and \( f'(x) \)
1. \( f(x) = 9x^2 + 4x - 1 \)
2. \( f'(x) = 18x + 4 \) (by differentiating \( f(x) \) with respect to \( x \)).

### Step 2: Use Newton’s Method for Each Initial Value

#### Case 1: \( x_0 = -1 \)
1. **Calculate \( f(x_0) \) and \( f'(x_0) \):**
   - \( f(-1) = 9(-1)^2 + 4(-1) - 1 = 9 - 4 - 1 = 4 \)
   - \( f'(-1) = 18(-1) + 4 = -18 + 4 = -14 \)

2. **Calculate \( x_1 \):**
   \[
   x_1 = x_0 - \frac{f(x_0)}{f'(x_0)} = -1 - \frac{4}{-14} = -1 + \frac{2}{7} = -\frac{5}{7} \approx -0.7143
   \]

3. **Calculate \( f(x_1) \) and \( f'(x_1) \):**
   - \( f\left(-\frac{5}{7}\right) = 9\left(-\frac{5}{7}\right)^2 + 4\left(-\frac{5}{7}\right) - 1 \approx 1.3061 \)
   - \( f'\left(-\frac{5}{7}\right) = 18\left(-\frac{5}{7}\right) + 4 \approx -8.8571 \)

4. **Calculate \( x_2 \):**
   \[
   x_2 = x_1 - \frac{f(x_1)}{f'(x_1)} \approx -0.7143 - \frac{1.3061}{-8.8571} \approx -0.5679
   \]

So, for \( x_0 = -1 \), we get \( x_2 \approx -0.5679 \).

#### Case 2: \( x_0 = 1 \)
1. **Calculate \( f(x_0) \) and \( f'(x_0) \):**
   - \( f(1) = 9(1)^2 + 4(1) - 1 = 9 + 4 - 1 = 12 \)
   - \( f'(1) = 18(1) + 4 = 18 + 4 = 22 \)

2. **Calculate \( x_1 \):**
   \[
   x_1 = x_0 - \frac{f(x_0)}{f'(x_0)} = 1 - \frac{12}{22} = 1 - \frac{6}{11} = \frac{5}{11} \approx 0.4545
   \]

3. **Calculate \( f(x_1) \) and \( f'(x_1) \):**
   - \( f\left(\frac{5}{11}\right) = 9\left(\frac{5}{11}\right)^2 + 4\left(\frac{5}{11}\right) - 1 \approx 3.0992 \)
   - \( f'\left(\frac{5}{11}\right) = 18\left(\frac{5}{11}\right) + 4 \approx 11.1818 \)

4. **Calculate \( x_2 \):**
   \[
   x_2 = x_1 - \frac{f(x_1)}{f'(x_1)} \approx 0.4545 - \frac{3.0992}{11.1818} \approx 0.1780
   \]

So, for \( x_0 = 1 \), we get \( x_2 \approx 0.1780 \).

### Summary
- For \( x_0 = -1 \): \( x_2 \approx -0.5679 \)
- For \( x_0 = 1 \): \( x_2 \approx 0.1780 \)

Would you like further clarification on any part of this solution? 

### Related Questions
1. How does Newton's Method converge to a solution?
2. Can Newton's Method fail to converge? If so, under what conditions?
3. How do initial guesses affect the convergence in Newton’s Method?
4. What is the geometric interpretation of Newton's Method?
5. How can we estimate the error in Newton’s Method at each iteration?

### Tip
Always check the derivative \( f'(x) \) in Newton's Method. If \( f'(x) \) is close to zero, Newton's Method may fail to converge or give inaccurate results.

Use Newton's method to estimate the solutions of the equation 9x^2 + 4x - 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 to estimate the solutions of 9x^2 + 4x - 1 = 0 using Newton's Method. This step-by-step guide explains how to find approximations for roots with initial guesses of x_0 = -1 and x_0 = 1, focusing on calculating x_2 in each case.

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