If `x^(2)+px+q=0` and `x^(2)+qx+p=0` have a common root, prove that either `p=q` or `1+p+q=0`.

Answer

To prove that either \( p = q \) or \( 1 + p + q = 0 \) given that the equations \( x^2 + px + q = 0 \) and \( x^2 + qx + p = 0 \) have a common root, we can follow these steps:

### Step 1: Let the common root be \( \alpha \).

Since \( \alpha \) is a common root, it satisfies both equations:
1. \( \alpha^2 + p\alpha + q = 0 \) (Equation 1)
2. \( \alpha^2 + q\alpha + p = 0 \) (Equation 2)

### Step 2: Set the equations equal to each other.

From both equations, we can express them as:
- From Equation 1: \( \alpha^2 = -p\alpha - q \)
- From Equation 2: \( \alpha^2 = -q\alpha - p \)

Since both expressions equal \( \alpha^2 \), we can set them equal to each other:
\[ -p\alpha - q = -q\alpha - p \]

### Step 3: Rearranging the equation.

Rearranging gives:
\[ -p\alpha + q\alpha = -p + q \]
\[ (q - p)\alpha = q - p \]

### Step 4: Analyze the results.

This gives us two cases:
1. If \( q - p \neq 0 \), we can divide both sides by \( q - p \):
\[ \alpha = 1 \]
2. If \( q - p = 0 \), then \( p = q \).

### Step 5: Substitute \( \alpha = 1 \) back into one of the original equations.

Substituting \( \alpha = 1 \) into Equation 1:
\[ 1^2 + p(1) + q = 0 \]
This simplifies to:
\[ 1 + p + q = 0 \]
Thus:
\[ 1 + p + q = 0 \]

### Conclusion

We have shown that either \( p = q \) or \( 1 + p + q = 0 \).