If $a+b+c=0$, $a\ne 0,ab,c\in Q$, then both the roots of the equation $a{x}^{2}+bx+c=0$ are
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Rational
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Non-real
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Irrational
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Zero
Given a+b+c=0, we can see that x=1 is a root of ax^2+bx+c=0 because a(1)^2+b(1)+c = a+b+c = 0. Since the product of the roots is c/a and one root is 1, the other root is c/a. Since a, b, c are rational, both roots are rational.
Because a, b, and c are rational numbers and a is not zero, the discriminant of the quadratic equation, calculated as D = b^2 - 4ac, will also be a rational number. We use the given condition a + b + c = 0 to rewrite the discriminant by substituting c = -(a + b). This gives D = b^2 - 4a(-(a + b)), which expands to D = b^2 + 4a^2 + 4ab. This expression factors perfectly into a perfect square, giving D = (2a + b)^2. Since the discriminant is the square of a rational number (2a + b), its square root is also rational. Consequently, the quadratic formula yields two real, rational roots for the equation.