If `b_(1)b_(2) = 2(c_(1) + c_(2))`, then at least one of the equations `x^(2) + b_(1)x + c_(1) = 0` and `x^(2) + b_(2)x + c_(2) = 0` has

Options

A. imaginary roots
B. real roots
C. purely imaginary roots
D. none of these

Explanation

To solve the problem, we need to analyze the given condition and the two quadratic equations. Let's break it down step by step.

### Step 1: Understand the Given Condition
We are given that:
\[ b_1 b_2 = 2(c_1 + c_2) \]

### Step 2: Rewrite the Condition
We can rewrite the condition as:
\[ \frac{b_1 b_2}{2} = c_1 + c_2 \]

### Step 3: Write the Quadratic Equations
We have two quadratic equations:
1. \( x^2 + b_1 x + c_1 = 0 \) (Equation 1)
2. \( x^2 + b_2 x + c_2 = 0 \) (Equation 2)

### Step 4: Find the Discriminants
The discriminant of a quadratic equation \( ax^2 + bx + c = 0 \) is given by:
\[ D = b^2 - 4ac \]

For Equation 1:
\[ D_1 = b_1^2 - 4c_1 \]

For Equation 2:
\[ D_2 = b_2^2 - 4c_2 \]

### Step 5: Add the Discriminants
Now, let's add both discriminants:
\[ D_1 + D_2 = (b_1^2 - 4c_1) + (b_2^2 - 4c_2) \]
\[ D_1 + D_2 = b_1^2 + b_2^2 - 4(c_1 + c_2) \]

### Step 6: Substitute the Condition
Substituting \( c_1 + c_2 \) from the rewritten condition:
\[ D_1 + D_2 = b_1^2 + b_2^2 - 4 \left( \frac{b_1 b_2}{2} \right) \]
\[ D_1 + D_2 = b_1^2 + b_2^2 - 2b_1 b_2 \]

### Step 7: Factor the Expression
The expression can be factored as:
\[ D_1 + D_2 = (b_1 - b_2)^2 \]

### Step 8: Analyze the Result
Since \( (b_1 - b_2)^2 \) is always non-negative, we have:
\[ D_1 + D_2 \geq 0 \]

### Step 9: Conclusion about the Roots
If \( D_1 + D_2 > 0 \), then at least one of \( D_1 \) or \( D_2 \) must be positive. This means at least one of the quadratic equations has real roots.

### Final Answer
Thus, at least one of the equations \( x^2 + b_1 x + c_1 = 0 \) or \( x^2 + b_2 x + c_2 = 0 \) has real roots.