If the equation `x^2-a_1x+b_1=0` and. `x^2-a_2x+b_2=0`, have one root in common and the second equation has equal roots, then which one of the following is a valid relation?

Options

A. `2(b_1-b_2)=a_1a_2`
B. `a_2(a_1+a_2)= b_1b_2`
C. `2(a_1-a_2)=b_1b_2`
D. `2(b_1+b_2)=a_1a_2`

Answer

To solve the problem, we need to analyze two quadratic equations:

1. \( x^2 - a_1x + b_1 = 0 \)
2. \( x^2 - a_2x + b_2 = 0 \)

Given that these equations share one common root and the second equation has equal roots, we will derive the necessary relationships step by step.

### Step 1: Identify the roots
Let the common root be \( \alpha \) and the other root of the first equation be \( \beta \). Thus, the roots of the first equation are \( \alpha \) and \( \beta \).

### Step 2: Apply the sum of roots for the first equation
From Vieta's formulas, the sum of the roots of the first equation is given by:
\[ \alpha + \beta = a_1 \]

### Step 3: Apply the sum of roots for the second equation
Since the second equation has equal roots, let both roots be \( \alpha \) (the common root). Thus, the sum of the roots for the second equation is:
\[ \alpha + \alpha = 2\alpha = a_2 \]

### Step 4: Apply the product of roots for the first equation
The product of the roots for the first equation is:
\[ \alpha \cdot \beta = b_1 \]

### Step 5: Apply the product of roots for the second equation
For the second equation, since both roots are \( \alpha \), we have:
\[ \alpha \cdot \alpha = \alpha^2 = b_2 \]

### Step 6: Use the condition for equal roots
The condition for the second equation to have equal roots is given by the discriminant:
\[ b^2 - 4ac = 0 \]
Substituting the values from the second equation:
\[ (-a_2)^2 - 4 \cdot 1 \cdot b_2 = 0 \]
This simplifies to:
\[ a_2^2 = 4b_2 \]

### Step 7: Substitute \( a_2 \) and \( b_2 \)
From step 3, we have \( a_2 = 2\alpha \), and from step 5, we have \( b_2 = \alpha^2 \). Substituting these into the equation from step 6 gives:
\[ (2\alpha)^2 = 4(\alpha^2) \]
This is consistent and confirms our expressions.

### Step 8: Relate \( b_1 \) and \( b_2 \)
From step 4, we have \( \alpha \cdot \beta = b_1 \) and \( \alpha^2 = b_2 \). Thus, we can express \( \beta \) in terms of \( b_1 \) and \( b_2 \):
\[ \beta = \frac{b_1}{\alpha} \]

### Step 9: Substitute \( \beta \) into the sum of roots equation
From step 2, we have:
\[ \alpha + \frac{b_1}{\alpha} = a_1 \]
Multiplying through by \( \alpha \) gives:
\[ \alpha^2 + b_1 = a_1 \alpha \]
Substituting \( b_2 = \alpha^2 \) into this equation results in:
\[ b_2 + b_1 = a_1 \alpha \]

### Step 10: Final relation
Using the earlier derived relationships, we can express \( a_1 \) in terms of \( a_2 \) and \( b_1, b_2 \):
\[ a_1 a_2 = 2b_1 + b_2 \]

### Conclusion
Thus, the valid relation is:
\[ a_1 a_2 = 2b_1 + b_2 \]