Let $x_{1}, x_{2}$ be the roots of $a x^{2} + b x + c = 0$ and $x_{1} . x_{2} < 0$ , $x_{1} + x_{2}$ is non - zero. Roots of $x_{1} (x - x_{2})^{2} + x_{2} (x - x_{1})^{2} = 0$ are

Real and of opposite signs

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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 A
Real and of opposite signs

Given $x_{1} (x - x_{2})^{2} + x_{2} (x - x_{1})^{2} = 0$
$x_{1} (x^{2} + x_{2}^{2} - 2 x x_{2}) + x_{2} (x^{2} + x_{1}^{2} - 2 x x_{1}) = 0$
$\Rightarrow x^{2} (x_{1} + x_{2}) - 4 x x_{1} x_{2} + x_{1} x_{2} (x_{1} + x_{2}) = 0 Δ = 16 x_{1}^{2} x_{2}^{2} - 4 x_{1} x_{2} (x_{1} + x_{2})^{2} > 0 (∵ x_{1} x_{2} < 0 g i v e n)$
$∴$ Real roots
Also, product of roots $= x_{1} x_{2} < 0$
$∴$ Roots are of opposite sign.

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