Determine the negative interval using false position method of x² - x - 6.

The False Position Method (or Regula Falsi) is a root-finding algorithm that combines the bisection method's bracketing with a linear interpolation approach. To apply the False Position Method to the function \( f(x) = x^2 - x - 6 \), we need to find an interval \([a, b]\) where the function changes sign, i.e., \( f(a) \cdot f(b) < 0 \).

### Steps for the False Position Method:

1. **Define the function:**
   \[
   f(x) = x^2 - x - 6
   \]

2. **Choose the initial interval \([a, b]\):**
   We need to find \(a\) and \(b\) such that \( f(a) \) and \( f(b) \) have opposite signs.

   Let's test the interval \([-3, -2]\):
   \[
   f(-3) = (-3)^2 - (-3) - 6 = 9 + 3 - 6 = 6
   \]
   \[
   f(-2) = (-2)^2 - (-2) - 6 = 4 + 2 - 6 = 0
   \]

   Let's test another interval \([-3, -1]\):
   \[
   f(-1) = (-1)^2 - (-1) - 6 = 1 + 1 - 6 = -4
   \]

   Since \( f(-3) \cdot f(-1) < 0 \), the interval \([-3, -1]\) contains a root.

3. **Apply the False Position Method:**

   The formula for the False Position Method is:
   \[
   c = b - \frac{f(b)(b - a)}{f(b) - f(a)}
   \]

   Using the interval \([-3, -1]\):
   \[
   a = -3, \quad b = -1
   \]
   \[
   f(a) = f(-3) = 6, \quad f(b) = f(-1) = -4
   \]

4. **Compute the first approximation \(c\):**
   \[
   c = b - \frac{f(b)(b - a)}{f(b) - f(a)} = -1 - \frac{-4(-1 + 3)}{-4 - 6}
   \]
   \[
   c = -1 - \frac{-4(2)}{-10} = -1 - \frac{-8}{-10} = -1 + 0.8 = -1.8
   \]

5. **Evaluate the function at the new point \(c\):**
   \[
   f(-1.8) = (-1.8)^2 - (-1.8) - 6 = 3.24 + 1.8 - 6 = -0.96
   \]

6. **Determine the new interval:**
   Since \( f(-1.8) \) and \( f(-1) \) have opposite signs, update the interval to \([-1.8, -1]\).

7. **Repeat the process:**

   1. Compute the new point \(c\):
      \[
      c = b - \frac{f(b)(b - a)}{f(b) - f(a)} = -1 - \frac{-4(-1 + 1.8)}{-4 + 0.96}
      \]
      \[
      c = -1 - \frac{-4(0.8)}{-3.04} = -1 - \frac{-3.2}{-3.04} = -1 + 1.0526 \approx -1.0526
      \]

   2. Evaluate the function at the new point \(c\):
      \[
      f(-1.0526) = (-1.0526)^2 - (-1.0526) - 6 = 1.10896576 + 1.0526 - 6 = -3.83843424
      \]

   3. Determine the new interval:
      Since \( f(-1.0526) \) and \( f(-1) \) have opposite signs, update the interval to \([-1.8, -1.0526]\).

Continue iterating until the interval length is less than a specified tolerance (e.g., 0.001).

### Summary of Iterations:
1. Interval: \([-3, -1]\)
2. First approximation \(c\): \(-1.8\)
3. Interval: \([-1.8, -1]\)
4. Second approximation \(c\): \(-1.0526\)
5. Interval: \([-1.8, -1.0526]\)

Do you want details on further iterations, or do you have any specific questions?

### Related Questions:
1. What is the main idea behind the False Position Method?
2. How do you determine the initial interval for the False Position Method?
3. What are the advantages of the False Position Method?
4. What are the limitations of the False Position Method?
5. How does the False Position Method compare to the Bisection Method?
6. Can the False Position Method be used for functions with multiple roots?
7. How does the convergence rate of the False Position Method compare to other root-finding methods?
8. What are some practical applications of the False Position Method in science and engineering?

### Tip:
In the False Position Method, the interval \([a, b]\) is updated based on the sign of the function at the new point \(c\), ensuring that the root is always bracketed within the interval.

Learn how to solve the quadratic equation x² - x - 6 using the False Position Method with a step-by-step explanation. This problem involves applying root-finding algorithms to find the roots of quadratic equations. Suitable for advanced high school students and anyone interested in numerical methods.

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