If the equations `x^2+px+q=0` and `x^2+qx+p=0` have a common root, which of the following can be the common root?

Options

A. 1
B. `-1`
C. `-2`
D. `(-1)/2`

Answer

To solve the problem, we need to find the common root of the two quadratic equations \(x^2 + px + q = 0\) and \(x^2 + qx + p = 0\).

### Step-by-Step Solution:

1. **Assume a Common Root**: Let \(\alpha\) be the common root of both equations. Therefore, we can write:
\[ \alpha^2 + p\alpha + q = 0 \quad \text{(1)} \]
\[ \alpha^2 + q\alpha + p = 0 \quad \text{(2)} \]

2. **Subtract the Two Equations**: Now, we subtract equation (2) from equation (1):
\[ (\alpha^2 + p\alpha + q) - (\alpha^2 + q\alpha + p) = 0 \]
This simplifies to:
\[ p\alpha - q\alpha + q - p = 0 \]
Which can be rearranged to:
\[ (p - q)\alpha + (q - p) = 0 \]

3. **Factor Out Common Terms**: We can factor out the common terms:
\[ (p - q)(\alpha - 1) = 0 \]

4. **Analyze the Factors**: From the equation \((p - q)(\alpha - 1) = 0\), we have two cases:
- Case 1: \(p - q = 0\) (This means \(p = q\))
- Case 2: \(\alpha - 1 = 0\) (This means \(\alpha = 1\))

5. **Conclusion**: Since the problem asks for a common root, and we found that if \(p \neq q\), then \(\alpha = 1\) is a common root. If \(p = q\), both equations become identical, and any root of the equation can be considered.

Thus, the common root can be \(1\) when \(p \neq q\).