if `x+1/x=P, then x^2+1/x^2=`

Options

A. `p^2`
B. `p^2+2`
C. `p^2-2`
D. `p^2-4`

Answer

To solve the problem, we need to find the value of \( x^2 + \frac{1}{x^2} \) given that \( x + \frac{1}{x} = P \).

### Step-by-Step Solution:

1. **Start with the given equation:**
\[ x + \frac{1}{x} = P \]

2. **Square both sides:**
\[ \left( x + \frac{1}{x} \right)^2 = P^2 \]

3. **Use the identity for squaring a binomial:**
The identity states that:
\[ (a + b)^2 = a^2 + b^2 + 2ab \]
Here, let \( a = x \) and \( b = \frac{1}{x} \). Therefore:
\[ \left( x + \frac{1}{x} \right)^2 = x^2 + \frac{1}{x^2} + 2 \cdot x \cdot \frac{1}{x} \]

4. **Simplify the expression:**
Since \( x \cdot \frac{1}{x} = 1 \):
\[ x^2 + \frac{1}{x^2} + 2 = P^2 \]

5. **Rearrange to find \( x^2 + \frac{1}{x^2} \):**
\[ x^2 + \frac{1}{x^2} = P^2 - 2 \]

### Final Answer:
\[ x^2 + \frac{1}{x^2} = P^2 - 2 \]