If `p != 0, q != 0` and the roots of `x^(2) + px +q = 0` are p and q, then (p, q) =

Options

A. (1, -2)
B. (1, 1)
C. (2, -2)
D. (-1, 2)

Answer

To solve the problem, we need to find the values of \( p \) and \( q \) given that the roots of the quadratic equation \( x^2 + px + q = 0 \) are \( p \) and \( q \).

### Step-by-Step Solution:

1. **Identify the Roots**: The roots of the quadratic equation \( x^2 + px + q = 0 \) are given as \( p \) and \( q \).

2. **Use the Sum of Roots**: According to Vieta's formulas, the sum of the roots \( p + q \) is equal to \( -\frac{\text{coefficient of } x}{\text{coefficient of } x^2} \). Thus,
\[ p + q = -p \quad \text{(Equation 1)} \]

3. **Use the Product of Roots**: Similarly, the product of the roots \( pq \) is equal to \( \frac{\text{constant term}}{\text{coefficient of } x^2} \). Therefore,
\[ pq = q \quad \text{(Equation 2)} \]

4. **Substituting Equation 1 into Equation 2**: From Equation 1, we can express \( q \) in terms of \( p \):
\[ q = -p - p = -2p \]
Now substitute this value of \( q \) into Equation 2:
\[ p(-2p) = -2p \]
Simplifying this gives:
\[ -2p^2 = -2p \]

5. **Rearranging the Equation**: We can rearrange the equation:
\[ -2p^2 + 2p = 0 \]
Factoring out \( -2p \):
\[ -2p(p - 1) = 0 \]

6. **Finding the Values of \( p \)**: This gives us two possible solutions:
\[ p = 0 \quad \text{or} \quad p = 1 \]
Since \( p \neq 0 \) (as given in the problem), we have:
\[ p = 1 \]

7. **Finding the Value of \( q \)**: Now substituting \( p = 1 \) back into the expression for \( q \):
\[ q = -2p = -2(1) = -2 \]

8. **Final Result**: Thus, the values of \( (p, q) \) are:
\[ (p, q) = (1, -2) \]

### Summary:
The final answer is \( (p, q) = (1, -2) \).