`If a, b, c, d in R` such that `a lt b lt c lt d,` then roots of the equation `(x-a)(x-c)+2(x-b)(x-d) = 0`

Options

A. are imaginary
B. are equal
C. are real and distinct lying between a and b
D. real and distinct lying between a and d.

Explanation

To solve the equation \((x-a)(x-c) + 2(x-b)(x-d) = 0\) given that \(a < b < c < d\), we will analyze the function step by step.

### Step 1: Define the function
Let \(f(x) = (x-a)(x-c) + 2(x-b)(x-d)\).

### Step 2: Evaluate \(f(a)\)
Substituting \(x = a\):
\[ f(a) = (a-a)(a-c) + 2(a-b)(a-d) = 0 + 2(a-b)(a-d) \]
Since \(a < b\) and \(a < d\), both \(a-b\) and \(a-d\) are negative. Thus, \(f(a) < 0\).

### Step 3: Evaluate \(f(b)\)
Substituting \(x = b\):
\[ f(b) = (b-a)(b-c) + 2(b-b)(b-d) = (b-a)(b-c) + 0 \]
Since \(b > a\) and \(b < c\), \(b-a > 0\) and \(b-c < 0\). Therefore, \(f(b) < 0\).

### Step 4: Evaluate \(f(c)\)
Substituting \(x = c\):
\[ f(c) = (c-a)(c-c) + 2(c-b)(c-d) = 0 + 2(c-b)(c-d) \]
Here, \(c > b\) and \(c < d\), so \(c-b > 0\) and \(c-d < 0\). Thus, \(f(c) < 0\).

### Step 5: Evaluate \(f(d)\)
Substituting \(x = d\):
\[ f(d) = (d-a)(d-c) + 2(d-b)(d-d) = (d-a)(d-c) + 0 \]
Since \(d > a\) and \(d > c\), both \(d-a > 0\) and \(d-c > 0\). Therefore, \(f(d) > 0\).

### Step 6: Analyze the sign changes
Now we have:
- \(f(a) < 0\)
- \(f(b) < 0\)
- \(f(c) < 0\)
- \(f(d) > 0\)

Since \(f(x)\) is continuous, by the Intermediate Value Theorem, there must be at least one root between \(c\) and \(d\) (where the function changes from negative to positive).

### Step 7: Conclusion about the roots
Since we have \(f(a) < 0\) and \(f(b) < 0\), and \(f(c) < 0\) but \(f(d) > 0\), we conclude that there are two roots:
1. One root lies between \(a\) and \(b\).
2. The second root lies between \(c\) and \(d\).

Thus, the roots of the equation \((x-a)(x-c) + 2(x-b)(x-d) = 0\) are real and distinct, lying between \(a\) and \(d\).

### Final Answer
The roots of the equation are real and distinct, lying between \(a\) and \(d\).