If `(x+2)` is a common factor of `(px^2+qx+r)` and `(qx^2+px+r)` then (a) ` p=q` or `p+q+r=0` (b)`p=r` or `p+q+r=0` (c) `q=r` or `p+q+r=0` (d)`p=q=-1/2r`

Options

A. `|d| le |D|`
B. `|d| ge|D|`
C. `|d| = |D|`
D. None of these

Explanation

To solve the problem, we need to determine the relationship between the coefficients \( p \), \( q \), and \( r \) given that \( (x + 2) \) is a common factor of the polynomials \( f(x) = px^2 + qx + r \) and \( g(x) = qx^2 + px + r \).

### Step-by-step Solution:

1. **Identify the Common Factor Condition**:
Since \( (x + 2) \) is a common factor, we know that \( f(-2) = 0 \) and \( g(-2) = 0 \).

2. **Calculate \( f(-2) \)**:
\[ f(-2) = p(-2)^2 + q(-2) + r = 4p - 2q + r \]
Setting this equal to zero gives us:
\[ 4p - 2q + r = 0 \quad \text{(1)} \]

3. **Calculate \( g(-2) \)**:
\[ g(-2) = q(-2)^2 + p(-2) + r = 4q - 2p + r \]
Setting this equal to zero gives us:
\[ 4q - 2p + r = 0 \quad \text{(2)} \]

4. **Equate the Two Expressions for \( r \)**:
From equation (1), we can express \( r \):
\[ r = -4p + 2q \quad \text{(3)} \]
From equation (2), we can express \( r \) as well:
\[ r = -4q + 2p \quad \text{(4)} \]

5. **Set the Two Expressions for \( r \) Equal**:
Equating equations (3) and (4):
\[ -4p + 2q = -4q + 2p \]

6. **Rearranging the Equation**:
Rearranging gives us:
\[ -4p + 4q = 2p - 2q \]
\[ 6q = 6p \]
Dividing both sides by 6:
\[ p = q \quad \text{(5)} \]

7. **Substituting \( p = q \) into \( r \)**:
Substitute \( p = q \) into equation (3):
\[ r = -4p + 2p = -2p \]
Therefore, we can express \( p \) and \( q \) in terms of \( r \):
\[ p = q = -\frac{1}{2}r \quad \text{(6)} \]

### Conclusion:
From our calculations, we find that \( p = q = -\frac{1}{2}r \).

### Answer:
The correct option is (d) \( p = q = -\frac{1}{2}r \).