The roots of `(1)/(x-1)+(1)/(x-2)+(1)/(x-3)=0` are :

Options

A. Real
B. Real and equal
C. Real and unequal
D. Imaginary

Answer

To solve the equation \(\frac{1}{x-1} + \frac{1}{x-2} + \frac{1}{x-3} = 0\), we will follow these steps:

### Step 1: Combine the fractions
To combine the fractions, we will find a common denominator, which is \((x-1)(x-2)(x-3)\).

The equation becomes:
\[ \frac{(x-2)(x-3) + (x-1)(x-3) + (x-1)(x-2)}{(x-1)(x-2)(x-3)} = 0 \]
Since the denominator cannot be zero, we focus on the numerator:
\[ (x-2)(x-3) + (x-1)(x-3) + (x-1)(x-2) = 0 \]

### Step 2: Expand the numerator
Now, we will expand each term in the numerator:
1. \((x-2)(x-3) = x^2 - 5x + 6\)
2. \((x-1)(x-3) = x^2 - 4x + 3\)
3. \((x-1)(x-2) = x^2 - 3x + 2\)

Adding these together gives:
\[ (x^2 - 5x + 6) + (x^2 - 4x + 3) + (x^2 - 3x + 2) = 0 \]
Combining like terms:
\[ 3x^2 - 12x + 11 = 0 \]

### Step 3: Form the quadratic equation
We now have the quadratic equation:
\[ 3x^2 - 12x + 11 = 0 \]

### Step 4: Calculate the discriminant
To determine the nature of the roots, we calculate the discriminant \(D\):
\[ D = b^2 - 4ac \]
Here, \(a = 3\), \(b = -12\), and \(c = 11\):
\[ D = (-12)^2 - 4 \cdot 3 \cdot 11 = 144 - 132 = 12 \]

### Step 5: Determine the nature of the roots
Since the discriminant \(D = 12\) is positive, this indicates that the roots of the quadratic equation are real and unequal.

### Final Answer
The roots of the equation \(\frac{1}{x-1} + \frac{1}{x-2} + \frac{1}{x-3} = 0\) are real and unequal.

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