If `a, b, c` are real and `a!=b`, then the roots ofthe equation, `2(a-b)x^2-11(a + b + c) x-3(a-b) = 0` are :

Options

A. real and equation
B. real and unequal
C. purely imaginary
D. none of these

Explanation

To solve the equation \(2(a-b)x^2 - 11(a+b+c)x - 3(a-b) = 0\) and determine the nature of its roots, we will follow these steps:

### Step 1: Identify the coefficients
The given quadratic equation can be compared to the standard form \(Ax^2 + Bx + C = 0\). Here, we identify:
- \(A = 2(a-b)\)
- \(B = -11(a+b+c)\)
- \(C = -3(a-b)\)

### Step 2: Calculate the discriminant
The nature of the roots of a quadratic equation is determined by the discriminant \(D\), given by the formula:
\[ D = B^2 - 4AC \]
Substituting the values of \(A\), \(B\), and \(C\) into the discriminant formula:
\[ D = [-11(a+b+c)]^2 - 4[2(a-b)][-3(a-b)] \]

### Step 3: Simplify the discriminant
Calculating \(D\):
\[ D = 121(a+b+c)^2 - 4 \cdot 2(a-b) \cdot (-3(a-b)) \]
\[ D = 121(a+b+c)^2 + 24(a-b)^2 \]

### Step 4: Analyze the discriminant
Since \(a\), \(b\), and \(c\) are real numbers and \(a \neq b\), we know:
- \((a-b)^2 > 0\) (because \(a-b \neq 0\))
- \((a+b+c)^2 \geq 0\) (since it is a square)

Thus, both terms \(121(a+b+c)^2\) and \(24(a-b)^2\) are non-negative. Therefore:
\[ D = 121(a+b+c)^2 + 24(a-b)^2 > 0 \]
This implies that \(D\) is strictly greater than zero.

### Step 5: Conclusion about the roots
Since the discriminant \(D > 0\), we conclude that the roots of the quadratic equation are real and unequal.

### Final Answer
The roots of the equation \(2(a-b)x^2 - 11(a+b+c)x - 3(a-b) = 0\) are real and unequal.

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