If a,b and c are real numbers then the roots of the equation `(x-a)(x-b)+(x-b)(x-c)+(x-c)(x-a)=0` are always

Options

A. real
B. imaginary
C. positive
D. negative

Explanation

To solve the equation \((x-a)(x-b) + (x-b)(x-c) + (x-c)(x-a) = 0\) 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.

\[ (x-a)(x-b) = x^2 - (a+b)x + ab \]
\[ (x-b)(x-c) = x^2 - (b+c)x + bc \]
\[ (x-c)(x-a) = x^2 - (c+a)x + ca \]

Now, we can combine these expansions:

\[ (x-a)(x-b) + (x-b)(x-c) + (x-c)(x-a) = (x^2 - (a+b)x + ab) + (x^2 - (b+c)x + bc) + (x^2 - (c+a)x + ca) \]

### Step 2: Combine like terms
Now, we combine all the \(x^2\) terms, \(x\) terms, and constant terms:

\[ = 3x^2 - [(a+b) + (b+c) + (c+a)]x + (ab + bc + ca) \]

The coefficient of \(x\) simplifies to:

\[ -(2a + 2b + 2c) = -2(a + b + c) \]

So the equation becomes:

\[ 3x^2 - 2(a+b+c)x + (ab + bc + ca) = 0 \]

### Step 3: Identify coefficients
From the standard form \(Ax^2 + Bx + C = 0\), we identify:

- \(A = 3\)
- \(B = -2(a+b+c)\)
- \(C = ab + bc + ca\)

### Step 4: Calculate the discriminant
The discriminant \(D\) of a quadratic equation is given by:

\[ D = B^2 - 4AC \]

Substituting the values we found:

\[ D = [-2(a+b+c)]^2 - 4 \cdot 3 \cdot (ab + bc + ca) \]

Calculating \(D\):

\[ D = 4(a+b+c)^2 - 12(ab + bc + ca) \]

### Step 5: Analyze the discriminant
The expression \(4(a+b+c)^2 - 12(ab + bc + ca)\) can be rewritten as:

\[ D = 4[(a+b+c)^2 - 3(ab + bc + ca)] \]

### Step 6: Recognize the nature of the discriminant
The term \((a+b+c)^2 - 3(ab + bc + ca)\) is always non-negative because it can be expressed as:

\[ \frac{1}{2}[(a-b)^2 + (b-c)^2 + (c-a)^2] \geq 0 \]

This implies that \(D \geq 0\).

### Conclusion
Since the discriminant \(D\) is always non-negative, the roots of the equation are always real numbers.

### Final Answer
The roots of the equation \((x-a)(x-b) + (x-b)(x-c) + (x-c)(x-a) = 0\) are always real numbers.

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