The root of `(x-a)(x-a-1)+(x-a-1)(x-a-2)+(x-a)(x-a-2)=0, a in R` are always

Options

A. real and equal
B. imaginary
C. real and unequal
D. rational and equal

Answer

To solve the equation \((x-a)(x-a-1)+(x-a-1)(x-a-2)+(x-a)(x-a-2)=0\) and determine the nature of its roots, we will follow these steps:

### Step 1: Expand the equation
We start by expanding each term in the equation.

1. **First term**:
\[ (x-a)(x-a-1) = x^2 - (2a+1)x + (a^2 + a) \]

2. **Second term**:
\[ (x-a-1)(x-a-2) = x^2 - (2a+3)x + (a^2 + 3a + 2) \]

3. **Third term**:
\[ (x-a)(x-a-2) = x^2 - (2a+2)x + (a^2 + 2a) \]

### Step 2: Combine all expanded terms
Now we combine all the expanded terms:
\[ (x-a)(x-a-1) + (x-a-1)(x-a-2) + (x-a)(x-a-2) = 0 \]
This gives us:
\[ \left( x^2 - (2a+1)x + (a^2 + a) \right) + \left( x^2 - (2a+3)x + (a^2 + 3a + 2) \right) + \left( x^2 - (2a+2)x + (a^2 + 2a) \right) = 0 \]

### Step 3: Combine like terms
Now we collect like terms:
- **\(x^2\) terms**: \(x^2 + x^2 + x^2 = 3x^2\)
- **\(x\) terms**:
\[ -(2a+1) - (2a+3) - (2a+2) = -6a - 6 \]
- **Constant terms**:
\[ (a^2 + a) + (a^2 + 3a + 2) + (a^2 + 2a) = 3a^2 + 6a + 2 \]

Putting it all together, we have:
\[ 3x^2 - (6a + 6)x + (3a^2 + 6a + 2) = 0 \]

### Step 4: Identify coefficients
From the quadratic equation \(ax^2 + bx + c = 0\), we identify:
- \(a = 3\)
- \(b = -(6a + 6)\)
- \(c = 3a^2 + 6a + 2\)

### Step 5: Calculate the discriminant
The discriminant \(D\) is given by:
\[ D = b^2 - 4ac \]
Substituting the values:
\[ D = (-(6a + 6))^2 - 4 \cdot 3 \cdot (3a^2 + 6a + 2) \]
Calculating \(D\):
\[ D = (6a + 6)^2 - 12(3a^2 + 6a + 2) \]
\[ = 36a^2 + 72a + 36 - (36a^2 + 72a + 24) \]
\[ = 36a^2 + 72a + 36 - 36a^2 - 72a - 24 \]
\[ = 12 \]

### Step 6: Determine the nature of the roots
Since the discriminant \(D = 12\) is greater than 0, the roots of the quadratic equation are real and unequal.

### Final Answer
The roots of the equation are always **real and unequal**.

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