7.	Find the root of x2-2=0 by using

a.	Newton Raphson method( )  where Xo=1.5

b.	Secant formula () Xo=1.45   X1=1.5

Let's solve the problem step by step for each method.

### (a) Newton-Raphson Method

The Newton-Raphson formula is:

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

Here, \( f(x) = x^2 - 2 \) and \( f'(x) = 2x \).

#### Iteration Steps:

1. **Start with \( x_0 = 1.5 \):**

   \[
   f(1.5) = (1.5)^2 - 2 = 0.25
   \]
   \[
   f'(1.5) = 2 \cdot 1.5 = 3
   \]
   \[
   x_1 = 1.5 - \frac{0.25}{3} = 1.5 - 0.0833 = 1.4167
   \]

2. **Next iteration (\( x_1 = 1.4167 \)):**

   \[
   f(1.4167) = (1.4167)^2 - 2 \approx 0.00694
   \]
   \[
   f'(1.4167) = 2 \cdot 1.4167 = 2.8334
   \]
   \[
   x_2 = 1.4167 - \frac{0.00694}{2.8334} \approx 1.4167 - 0.00245 = 1.4143
   \]

3. **Next iteration (\( x_2 = 1.4143 \)):**

   \[
   f(1.4143) = (1.4143)^2 - 2 \approx 0.000007
   \]
   \[
   f'(1.4143) = 2 \cdot 1.4143 = 2.8286
   \]
   \[
   x_3 = 1.4143 - \frac{0.000007}{2.8286} \approx 1.4142
   \]

The root converges to **\( x \approx 1.4142 \)** after sufficient iterations.

---

### (b) Secant Method

The Secant formula is:

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

Given \( x_0 = 1.45 \) and \( x_1 = 1.5 \):

#### Iteration Steps:

1. **Compute \( f(x_0) \) and \( f(x_1) \):**
   \[
   f(1.45) = (1.45)^2 - 2 = 0.1025
   \]
   \[
   f(1.5) = (1.5)^2 - 2 = 0.25
   \]

2. **Find \( x_2 \):**
   \[
   x_2 = 1.5 - 0.25 \cdot \frac{1.5 - 1.45}{0.25 - 0.1025} = 1.5 - 0.25 \cdot \frac{0.05}{0.1475}
   \]
   \[
   x_2 = 1.5 - 0.0847 \approx 1.4153
   \]

3. **Compute \( f(x_2) \):**
   \[
   f(1.4153) = (1.4153)^2 - 2 \approx 0.00156
   \]

4. **Find \( x_3 \):**
   \[
   x_3 = 1.4153 - 0.00156 \cdot \frac{1.4153 - 1.5}{0.00156 - 0.25} \approx 1.4142
   \]

The root converges to **\( x \approx 1.4142 \)** after sufficient iterations.

---

### Final Results:
1. Newton-Raphson root: \( \approx 1.4142 \)
2. Secant method root: \( \approx 1.4142 \)

---

Would you like a detailed explanation of any part or need further assistance?  

Here are 5 related questions to explore:
1. Why does the Newton-Raphson method converge faster than the Secant method?
2. What are the conditions for convergence of these iterative methods?
3. How can the initial guesses influence the convergence of the Secant method?
4. What happens if the derivative in the Newton-Raphson method is zero at some point?
5. Can you apply these methods to find the root of a cubic polynomial?

**Tip:** Always check if the function and its derivative behave well around the initial guesses to avoid divergence in iterative methods.

7. Find the root of x^2 - 2 = 0 by using
a. Newton Raphson method where X0 = 1.5
b. Secant formula X0 = 1.45, X1 = 1.5

Learn how to find the root of x^2 - 2 = 0 using the Newton-Raphson method (X0 = 1.5) and Secant formula (X0 = 1.45, X1 = 1.5). This step-by-step solution covers numerical methods, including iteration processes and convergence, ideal for undergraduate-level students.

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