If the two equation `x^(2)-cx+d=0` and `x^(2)-ax+b=0` have one common root and the second has equal roots then 2(b + d) is equal to

Options

A. 0
B. a + c
C. ac
D. `-ac`

Answer

To solve the problem, we need to analyze the two quadratic equations given and use the properties of their roots. Let's denote the equations as follows:

1. \( x^2 - cx + d = 0 \) (Equation 1)
2. \( x^2 - ax + b = 0 \) (Equation 2)

### Step 1: Identify the common root
Let the common root be \( \alpha \). Since Equation 1 has roots \( \alpha \) and \( \beta \), we can express the relationships between the roots and coefficients using Vieta's formulas.

From Equation 1:
- Sum of roots: \( \alpha + \beta = c \) (1)
- Product of roots: \( \alpha \beta = d \) (2)

### Step 2: Analyze the second equation
Since Equation 2 has equal roots, let both roots be \( \alpha \). Thus, we have:
- Sum of roots: \( \alpha + \alpha = 2\alpha = a \) (3)
- Product of roots: \( \alpha \cdot \alpha = \alpha^2 = b \) (4)

### Step 3: Substitute and simplify
From equations (3) and (4), we can express \( a \) and \( b \) in terms of \( \alpha \):
- \( a = 2\alpha \)
- \( b = \alpha^2 \)

### Step 4: Substitute \( b \) and \( d \) into the expression \( 2(b + d) \)
We need to find \( 2(b + d) \):
- Substitute \( b \) and \( d \):
\[ 2(b + d) = 2(\alpha^2 + \alpha \beta) \]

### Step 5: Relate \( \beta \) to \( \alpha \)
From equation (1), we have \( \beta = c - \alpha \). Substitute this into the expression:
- \( d = \alpha \beta = \alpha (c - \alpha) = \alpha c - \alpha^2 \)

Now substituting \( d \) back into the expression:
\[ 2(b + d) = 2(\alpha^2 + (\alpha c - \alpha^2)) = 2(\alpha c) \]

### Step 6: Final expression
Thus, we have:
\[ 2(b + d) = 2\alpha c \]

### Conclusion
We conclude that the value of \( 2(b + d) \) is equal to \( 2\alpha c \).