Find the positive and negative roots of  x²-0.25x-0.375 = 0 using newton-raphson method.

To find the roots of the equation \( x^2 - 0.25x - 0.375 = 0 \) using the Newton-Raphson method, follow these steps:

### Step 1: Define the function and its derivative

Let \( f(x) = x^2 - 0.25x - 0.375 \).

The derivative of \( f(x) \) is:
\[ f'(x) = 2x - 0.25 \]

### Step 2: Choose initial guesses for the roots

To find the roots using the Newton-Raphson method, we need to choose initial guesses. Let's denote them by \( x_0 \).

### Step 3: Apply the Newton-Raphson iteration formula

The iteration formula for the Newton-Raphson method is:
\[ x_{n+1} = x_n - \frac{f(x_n)}{f'(x_n)} \]

We will iterate this formula until we converge to the roots.

### Finding the positive root

Let's start with an initial guess for the positive root, \( x_0 = 1 \).

1. **Iteration 1:**
   \[ x_1 = x_0 - \frac{f(x_0)}{f'(x_0)} \]
   \[ x_1 = 1 - \frac{1^2 - 0.25 \cdot 1 - 0.375}{2 \cdot 1 - 0.25} \]
   \[ x_1 = 1 - \frac{1 - 0.25 - 0.375}{2 - 0.25} \]
   \[ x_1 = 1 - \frac{0.375}{1.75} \]
   \[ x_1 = 1 - 0.2143 \]
   \[ x_1 \approx 0.7857 \]

2. **Iteration 2:**
   \[ x_2 = x_1 - \frac{f(x_1)}{f'(x_1)} \]
   \[ x_2 = 0.7857 - \frac{0.7857^2 - 0.25 \cdot 0.7857 - 0.375}{2 \cdot 0.7857 - 0.25} \]
   \[ x_2 \approx 0.7076 \]

3. **Iteration 3:**
   \[ x_3 = x_2 - \frac{f(x_2)}{f'(x_2)} \]
   \[ x_3 \approx 0.6956 \]

4. **Iteration 4:**
   \[ x_4 = x_3 - \frac{f(x_3)}{f'(x_3)} \]
   \[ x_4 \approx 0.6946 \]

### Finding the negative root

Let's start with an initial guess for the negative root, \( x_0 = -1 \).

1. **Iteration 1:**
   \[ x_1 = x_0 - \frac{f(x_0)}{f'(x_0)} \]
   \[ x_1 = -1 - \frac{(-1)^2 - 0.25 \cdot (-1) - 0.375}{2 \cdot (-1) - 0.25} \]
   \[ x_1 = -1 - \frac{1 + 0.25 - 0.375}{-2.25} \]
   \[ x_1 = -1 - \frac{0.875}{-2.25} \]
   \[ x_1 = -1 + 0.3889 \]
   \[ x_1 \approx -0.6111 \]

2. **Iteration 2:**
   \[ x_2 = x_1 - \frac{f(x_1)}{f'(x_1)} \]
   \[ x_2 = -0.6111 - \frac{(-0.6111)^2 - 0.25 \cdot (-0.6111) - 0.375}{2 \cdot (-0.6111) - 0.25} \]
   \[ x_2 \approx -0.6942 \]

3. **Iteration 3:**
   \[ x_3 = x_2 - \frac{f(x_2)}{f'(x_2)} \]
   \[ x_3 \approx -0.6956 \]

4. **Iteration 4:**
   \[ x_4 = x_3 - \frac{f(x_3)}{f'(x_3)} \]
   \[ x_4 \approx -0.6946 \]

### Conclusion

After several iterations, we can see that both roots converge to approximately:

\[ x_+ \approx 0.6946 \]
\[ x_- \approx -0.6946 \]

Would you like more details on any of these steps or have any questions?

### Relative Questions:

1. What are the advantages of the Newton-Raphson method compared to other root-finding algorithms?
2. How does the choice of the initial guess affect the convergence of the Newton-Raphson method?
3. Can the Newton-Raphson method fail to converge, and if so, under what conditions?
4. How can you verify the accuracy of the roots found using the Newton-Raphson method?
5. What are some common applications of the Newton-Raphson method in real-world problems?
6. How does the Newton-Raphson method handle functions with multiple roots or very close roots?
7. What modifications can be made to the Newton-Raphson method to improve its convergence rate?
8. How does the Newton-Raphson method compare to other numerical methods like the secant method or bisection method?

### Tip:
Always check the derivative of the function before using the Newton-Raphson method to ensure it does not become zero or very small, as this can cause the method to fail or converge slowly.

Discover how to find the positive and negative roots of the quadratic equation x² - 0.25x - 0.375 using the Newton-Raphson method. This problem involves applying iterative calculations with the Newton-Raphson formula. Suitable for advanced high school students and those interested in numerical methods.

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