$x_1$ and $x_2$ are the real roots of $ax^2+bx+c=0$ and $x_1x_2 < 0$. The roots of $x_1(x-x_2)^2+x_2(x-x_1)^2=0$ are

$ax^{ 2 }+bx+c=0$

Therefore, ${ x }_{ 1 }+{ x }_{ 2 }=-\dfrac { b }{ a }$ and ${ x }_{ 1 }{ x }_{ 2 }=\dfrac { c }{ a }$

and $\quad D={ b }^{ 2 }-4ac>0$

now, $x_{ 1 }(x-x_{ 2 })^{ 2 }+x_{ 2 }(x-x_{ 1 })^{ 2 }=0$

$\Rightarrow \left( { x }_{ 1 }+{ x }_{ 2 } \right) { x }^{ 2 }-4{ x }_{ 1 }{ x }_{ 2 }x+{ x }_{ 1 }{ x }_{ 2 }\left( { x }_{ 1 }+{ x }_{ 2 } \right) =0$

$\Rightarrow -\dfrac { b{ x }^{ 2 } }{ a } -\dfrac { 4cx }{ a } -\dfrac { bc }{ { a }^{ 2 } } =0$

$\Rightarrow ab{ x }^{ 2 }+4acx+bc=0$

$D_1=16a^2c^2-4b^2ac$

Since, ${x}_{1}{x}_{2}<0$ $ac<0$

Therefore, ${D}_{1}>0$

Now product of the roots of the equation $=\dfrac { bc }{ ab } ={ x }_{ 1 }{ x }_{ 2 }<0$

Therefore, roots are of opposite sign.

Ans: A