The set of values of `alpha` for which the quadratic equation
`(alpha + 2) x^(2) - 2 alpha x - alpha = 0`
has two roots on the number line symmetrically placed about the point 1 is

Options

A. {-1, 0}
B. {0,2}
C. `phi`
D. {0,1}

Answer

To solve the problem, we need to find the set of values of \( \alpha \) for which the quadratic equation

\[ (\alpha + 2)x^2 - 2\alpha x - \alpha = 0 \]

has two roots that are symmetrically placed about the point \( x = 1 \).

### Step 1: Understand the condition for symmetry
For the roots of the quadratic equation to be symmetrically placed about \( x = 1 \), if we denote the roots by \( r_1 \) and \( r_2 \), then we must have:

\[ r_1 + r_2 = 2 \cdot 1 = 2 \]

### Step 2: Use Vieta's formulas
According to Vieta's formulas, for a quadratic equation of the form \( ax^2 + bx + c = 0 \), the sum of the roots \( r_1 + r_2 = -\frac{b}{a} \). Here, we can identify:

- \( a = \alpha + 2 \)
- \( b = -2\alpha \)
- \( c = -\alpha \)

Thus, the sum of the roots is:

\[ r_1 + r_2 = -\frac{-2\alpha}{\alpha + 2} = \frac{2\alpha}{\alpha + 2} \]

### Step 3: Set the sum equal to 2
We set the sum of the roots equal to 2:

\[ \frac{2\alpha}{\alpha + 2} = 2 \]

### Step 4: Solve for \( \alpha \)
Cross-multiplying gives:

\[ 2\alpha = 2(\alpha + 2) \]

Expanding the right side:

\[ 2\alpha = 2\alpha + 4 \]

Subtracting \( 2\alpha \) from both sides:

\[ 0 = 4 \]

This equation is a contradiction, which means there are no values of \( \alpha \) that satisfy the condition for the roots to be symmetrically placed about \( x = 1 \).

### Final Answer
Thus, the set of values of \( \alpha \) for which the quadratic equation has two roots symmetrically placed about the point 1 is:

\[ \text{No values of } \alpha \]