Statement-I: If a + b + c = 0 and a, b, c are | vedclass

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(A) Given $a + b + c = 0$,we have $b + c = -a$,$c + a = -b$,and $a + b = -c$.Substituting these into the equation $(b + c - a)x^2 + (c + a - b)x + (a

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Statement-$I$ is true,Statement-$II$ is true,Statement-$II$ is the correct explanation of Statement-$I$.

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(A) Given $a + b + c = 0$,we have $b + c = -a$,$c + a = -b$,and $a + b = -c$.Substituting these into the equation $(b + c - a)x^2 + (c + a - b)x + (a + b - c) = 0$:$(-a - a)x^2 + (-b - b)x + (-c - c) = 0$$-2ax^2 - 2bx - 2c = 0$$ax^2 + bx + c = 0$.The discriminant $D = b^2 - 4ac$.Since $c = -(a + b)$,$D = b^2 - 4a(-(a + b)) = b^2 + 4a^2 + 4ab = (2a + b)^2$.Since $a, b, c$ are rational,$D$ is a perfect square of a rational number,so the roots are rational.Thus,Statement-$I$ is true.Statement-$II$ is also true because the discriminant is $(2a + b)^2$,which is a perfect square,and this explains why the roots are rational.

Statement-$I$: If $a + b + c = 0$ and $a, b, c$ are rational,then the roots of the equation $(b + c - a)x^2 + (c + a - b)x + (a + b - c) = 0$ are rational.
Statement-$II$: The discriminant of $(b + c - a)x^2 + (c + a - b)x + (a + b - c) = 0$ is a perfect square.

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Statement-$I$: If $a + b + c = 0$ and $a, b, c$ are rational,then the roots of the equation $(b + c - a)x^2 + (c + a - b)x + (a + b - c) = 0$ are rational.Statement-$II$: The discriminant of $(b + c - a)x^2 + (c + a - b)x + (a + b - c) = 0$ is a perfect square.