The roots of the equation $(b+c)x^2-(a+b+c)x+a=0 ::: (a,b,c :\epsilon:Q,b+c \neq a)$ are:
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Irrational and different
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Rational and different
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Imaginary and different
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Real and equal
The discriminant D = (a+b+c)^2 - 4(b+c)a = a^2 + b^2 + c^2 + 2ab + 2bc + 2ca - 4ab - 4ac = a^2 + b^2 + c^2 - 2ab - 2ac + 2bc = (a-b-c)^2 + 4bc. Since a,b,c are rational, the discriminant is a square of a rational number, so the roots are rational.
Compute the discriminant of the equation by evaluating negative (a plus b plus c) squared minus 4 times (b plus c) times a. This expands to a squared plus b squared plus c squared plus 2ab plus 2bc plus 2ca minus 4ab minus 4ac minus 4bc, which simplifies to a squared plus b squared plus c squared minus 2ab minus 2bc minus 2ca. By inspection, the quadratic expression x equals 1 satisfies the equation, and since the sum of roots is (a plus b plus c) divided by (b plus c), the other root must be a divided by (b plus c). Since a, b, and c are rational numbers, both roots are rational and different.