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

Options

A. p=1
B. p=1 or 0
C. p=-2
D. p=-2 or 0

Answer

To solve the problem, we start with the quadratic equation given:

\[ x^2 + px + q = 0 \]

where \( p \) and \( q \) are the roots of the equation.

### Step 1: 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 \( \alpha + \beta = -\frac{b}{a} \)
- The product of the roots \( \alpha \beta = \frac{c}{a} \)

In our case:
- The roots are \( p \) and \( q \)
- The coefficients are \( a = 1 \), \( b = p \), and \( c = q \)

Thus, we can write:
1. \( p + q = -\frac{p}{1} = -p \)
2. \( pq = \frac{q}{1} = q \)

### Step 2: Rearranging the Equations
From the first equation \( p + q = -p \), we can rearrange it:
\[ p + q + p = 0 \]
\[ 2p + q = 0 \]
This gives us our first equation:
\[ q = -2p \] (Equation 1)

From the second equation \( pq = q \), we can rearrange it:
\[ pq - q = 0 \]
Factoring out \( q \):
\[ q(p - 1) = 0 \]

### Step 3: Analyzing the Factors
From the factored equation \( q(p - 1) = 0 \), we have two cases:
1. \( q = 0 \)
2. \( p - 1 = 0 \) which gives \( p = 1 \)

### Step 4: Case 1: If \( q = 0 \)
If \( q = 0 \), substitute \( q \) back into Equation 1:
\[ 0 = -2p \]
This implies:
\[ p = 0 \]

### Step 5: Case 2: If \( p = 1 \)
If \( p = 1 \), substitute \( p \) back into Equation 1:
\[ q = -2(1) = -2 \]

### Conclusion
The possible values for \( p \) are:
- From Case 1: \( p = 0 \)
- From Case 2: \( p = 1 \)

Thus, the values of \( p \) can be either \( 0 \) or \( 1 \).

### Final Answer
The values of \( p \) are \( 0 \) and \( 1 \).

---