If the equations `x^(2) - px + q = 0 and x^(2) - ax + b = 0` have common root and the roots of the second equation are equal, then which one of the following is correct ?

Options

A. aq = 2 (b + q)
B. aq = (b + q)
C. ap = (b + q)
D. ap = b + q

Answer

To solve the problem, we need to analyze the given quadratic equations and their properties based on the conditions provided.

### Step-by-Step Solution:

1. **Identify the Equations:**
We have two quadratic equations:
\[ x^2 - px + q = 0 \quad \text{(Equation 1)} \]
\[ x^2 - ax + b = 0 \quad \text{(Equation 2)} \]

2. **Common Root:**
Let the common root be denoted as \(\alpha\). Since \(\alpha\) is a root of both equations, we can express this as:
\[ \alpha^2 - p\alpha + q = 0 \quad \text{(1)} \]
\[ \alpha^2 - a\alpha + b = 0 \quad \text{(2)} \]

3. **Equal Roots in Equation 2:**
The roots of Equation 2 are equal, which means the discriminant must be zero. The discriminant \(\Delta\) for Equation 2 is given by:
\[ \Delta = a^2 - 4b = 0 \]
This implies:
\[ a^2 = 4b \quad \text{(3)} \]

4. **Substituting \(\alpha\) in Both Equations:**
From Equation (1):
\[ \alpha^2 = p\alpha - q \]
From Equation (2):
\[ \alpha^2 = a\alpha - b \]

5. **Equating the Two Expressions for \(\alpha^2\):**
Since both expressions equal \(\alpha^2\), we can set them equal to each other:
\[ p\alpha - q = a\alpha - b \]
Rearranging gives:
\[ (p - a)\alpha = q - b \quad \text{(4)} \]

6. **Substituting \(b\) from Equation (3):**
From (3), we know that \(b = \frac{a^2}{4}\). Substituting this into (4):
\[ (p - a)\alpha = q - \frac{a^2}{4} \]

7. **Finding a Relation:**
Rearranging gives:
\[ (p - a)\alpha + \frac{a^2}{4} = q \]

8. **Final Relation:**
We can express this as:
\[ 4(p - a)\alpha + a^2 = 4q \]
This leads us to the final relation:
\[ 2b + q = ap \]

### Conclusion:
Thus, the correct relationship derived from the given conditions is:
\[ 2b + q = ap \]