If ` alpha ` is a positive root of the equation `x^4 +x^3 -4x^2 +x+1=0` then ` alpha +1/alpha `

Options

A. 2
B. `-2`
C. `3`
D. `4`

Answer

To solve the problem, we need to find the value of \( \alpha + \frac{1}{\alpha} \) given that \( \alpha \) is a positive root of the equation \( x^4 + x^3 - 4x^2 + x + 1 = 0 \).

### Step-by-Step Solution:

1. **Identify the polynomial**:
We have the polynomial \( f(x) = x^4 + x^3 - 4x^2 + x + 1 \).

2. **Factor the polynomial**:
We can try to factor the polynomial. We can rewrite it as:
\[ f(x) = x^4 + x^3 - 4x^2 + x + 1 = (x^2 - 1)^2 + x(x - 1)^2 \]
This suggests that \( f(x) \) can be factored further.

3. **Set up the factorization**:
We can factor \( f(x) \) as:
\[ f(x) = (x - 1)^2 (x^2 + 3x + 1) \]
This gives us two parts to analyze: \( (x - 1)^2 = 0 \) and \( x^2 + 3x + 1 = 0 \).

4. **Find the roots**:
From \( (x - 1)^2 = 0 \), we find \( x = 1 \) (with multiplicity 2).
Now we solve \( x^2 + 3x + 1 = 0 \) using the quadratic formula:
\[ x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} = \frac{-3 \pm \sqrt{3^2 - 4 \cdot 1 \cdot 1}}{2 \cdot 1} = \frac{-3 \pm \sqrt{5}}{2} \]
The roots are:
\[ x_1 = \frac{-3 + \sqrt{5}}{2}, \quad x_2 = \frac{-3 - \sqrt{5}}{2} \]

5. **Identify the positive root**:
Since \( \alpha \) is a positive root, we have:
\[ \alpha = 1 \quad \text{(the only positive root)} \]

6. **Calculate \( \alpha + \frac{1}{\alpha} \)**:
Now we can calculate:
\[ \alpha + \frac{1}{\alpha} = 1 + \frac{1}{1} = 1 + 1 = 2 \]

### Final Answer:
Thus, the value of \( \alpha + \frac{1}{\alpha} \) is \( \boxed{2} \).