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
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Rational
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Irrational
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Imaginary
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Equal
If a+b+c=0, then b+c = -a, c+a = -b, a+b = -c. The equation becomes (-a-a)x^2 + (-b-b)x + (-c-c) = 0 => -2ax^2 - 2bx - 2c = 0 => ax^2 + bx + c = 0. Since a, b, c are rational, the roots are rational if the discriminant b^2 - 4ac is a perfect square. However, the question implies general properties; given a+b+c=0, the roots are rational.
Given a + b + c = 0, we substitute c = -a - b into the coefficients of the equation (b + c - a)x^2 + (c + a - b)x + (a + b - c) = 0. This simplifies the coefficients to -2a, -2b, and 2a + 2b respectively, yielding the equation -2ax^2 - 2bx + (2a + 2b) = 0. The discriminant is D = (-2b)^2 - 4(-2a)(2a + 2b) = 4b^2 + 16a^2 + 16ab = 4(2a + b)^2. Since the discriminant is a perfect square and the coefficients are rational, the roots of the equation are rational.