Let `x_(1)` and `x_(2)` are the roots of `ax^(2)+bx+c=0 (a,b,c epsilon R) and x_(1).x_(2)lt0, x_(1)+x_(2)` is non zero, then the roots of `x_(1)(x-x_(2))^(2)+x_(2)(x-x_(1))^(2)= 0` are

Options

A. negative
B. real and opposite in sign
C. positive
D. non real

Explanation

To solve the given problem step by step, we start with the equation:

\[ x_1(x - x_2)^2 + x_2(x - x_1)^2 = 0 \]

### Step 1: Expand the equation
We can expand both terms in the equation:

1. Expand \( (x - x_2)^2 \):
\[ (x - x_2)^2 = x^2 - 2x x_2 + x_2^2 \]
Therefore,
\[ x_1(x - x_2)^2 = x_1(x^2 - 2x x_2 + x_2^2) = x_1 x^2 - 2 x_1 x x_2 + x_1 x_2^2 \]

2. Expand \( (x - x_1)^2 \):
\[ (x - x_1)^2 = x^2 - 2x x_1 + x_1^2 \]
Therefore,
\[ x_2(x - x_1)^2 = x_2(x^2 - 2x x_1 + x_1^2) = x_2 x^2 - 2 x_2 x x_1 + x_2 x_1^2 \]

Combining these expansions gives:
\[ x_1 x^2 - 2 x_1 x x_2 + x_1 x_2^2 + x_2 x^2 - 2 x_2 x x_1 + x_2 x_1^2 = 0 \]

### Step 2: Combine like terms
Now, we combine the terms:
\[ (x_1 + x_2)x^2 - 2(x_1 x_2 + x_2 x_1)x + (x_1 x_2^2 + x_2 x_1^2) = 0 \]
This simplifies to:
\[ (x_1 + x_2)x^2 - 4x_1 x_2 x + (x_1 x_2^2 + x_2 x_1^2) = 0 \]

### Step 3: Identify coefficients
Let:
- \( A = x_1 + x_2 \)
- \( B = -4 x_1 x_2 \)
- \( C = x_1 x_2 (x_1 + x_2) \)

Thus, we can rewrite the equation as:
\[ Ax^2 + Bx + C = 0 \]

### Step 4: Calculate the discriminant
The discriminant \( D \) of a quadratic equation \( Ax^2 + Bx + C = 0 \) is given by:
\[ D = B^2 - 4AC \]
Substituting our values:
\[ D = (-4 x_1 x_2)^2 - 4(x_1 + x_2)(x_1 x_2 (x_1 + x_2)) \]
This simplifies to:
\[ D = 16 x_1^2 x_2^2 - 4(x_1 + x_2)^2 (x_1 x_2) \]

### Step 5: Analyze the discriminant
Given that \( x_1 x_2 < 0 \) (the product of the roots is negative), we know:
- \( 16 x_1^2 x_2^2 \) is positive.
- The term \( -4(x_1 + x_2)^2 (x_1 x_2) \) is also positive because \( (x_1 + x_2)^2 \) is non-zero and \( x_1 x_2 < 0 \).

Thus, \( D > 0 \).

### Step 6: Conclusion
Since \( D > 0 \), the roots of the quadratic equation are real and distinct. Given that \( x_1 x_2 < 0 \), the roots must also be opposite in sign.

### Final Answer
The roots of the equation \( x_1(x - x_2)^2 + x_2(x - x_1)^2 = 0 \) are real and opposite in sign.

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