If (1 - p) is a root of quadratic equation `x^(2) + px + (1- p)=0`, then its roots are

Options

A. 0,1
B. -1, 2
C. 0, -1
D. -1, 1

Answer

To solve the problem, we need to find the roots of the quadratic equation given that \( (1 - p) \) is a root of the equation \( x^2 + px + (1 - p) = 0 \).

### Step-by-step Solution:

1. **Substituting the Root into the Equation:**
Since \( (1 - p) \) is a root of the equation, we can substitute \( x = 1 - p \) into the quadratic equation:
\[ (1 - p)^2 + p(1 - p) + (1 - p) = 0 \]

2. **Expanding the Equation:**
Now, let's expand the equation:
\[ (1 - p)^2 = 1 - 2p + p^2 \]
\[ p(1 - p) = p - p^2 \]
Therefore, substituting these into the equation gives:
\[ 1 - 2p + p^2 + p - p^2 + 1 - p = 0 \]

3. **Simplifying the Equation:**
Combine like terms:
\[ 1 + 1 - 2p + p - p + p^2 - p^2 = 0 \]
This simplifies to:
\[ 2 - 2p = 0 \]

4. **Solving for \( p \):**
Rearranging gives:
\[ 2 = 2p \implies p = 1 \]

5. **Finding the Roots of the Quadratic Equation:**
Now, substitute \( p = 1 \) back into the original quadratic equation:
\[ x^2 + 1x + (1 - 1) = 0 \implies x^2 + x = 0 \]

6. **Factoring the Quadratic:**
Factor the equation:
\[ x(x + 1) = 0 \]

7. **Finding the Roots:**
Setting each factor to zero gives:
\[ x = 0 \quad \text{or} \quad x + 1 = 0 \implies x = -1 \]

### Final Answer:
The roots of the quadratic equation are \( x = 0 \) and \( x = -1 \).