$x_{1}$ and $x_{2}$ are the real roots of $a x^{2} + b x + c = 0$ and $x_{1} x_{2} < 0$ . The roots of $x_{1} (x - x_{2})^{2} + x_{2} (x - x_{1})^{2} = 0$ are

real and of opposite sign

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negative

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positive

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non-real

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Solution

The correct option is C real and of opposite sign
$a x^{2} + b x + c = 0$
Therefore, $x_{1} + x_{2} = - \frac{b}{a}$ and $x_{1} x_{2} = \frac{c}{a}$
and $D = b^{2} - 4 a c > 0$
now, $x_{1} (x - x_{2})^{2} + x_{2} (x - x_{1})^{2} = 0$
$\Rightarrow (x_{1} + x_{2}) x^{2} - 4 x_{1} x_{2} x + x_{1} x_{2} (x_{1} + x_{2}) = 0$
$\Rightarrow - \frac{b x^{2}}{a} - \frac{4 c x}{a} - \frac{b c}{a^{2}} = 0$
$\Rightarrow a b x^{2} + 4 a c x + b c = 0$
$D_{1} = 16 a^{2} c^{2} - 4 b^{2} a c$
Since, $x_{1} x_{2} < 0$ $a c < 0$
Therefore, $D_{1} > 0$
Now product of the roots of the equation $= \frac{b c}{a b} = x_{1} x_{2} < 0$
Therefore, roots are of opposite sign.
Ans: A

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