If a + b + c = 0,a ne 0,and a, b, c in mathbb | vedclass

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(A) Given the quadratic equation $ax^2 + bx + c = 0$ where $a, b, c \in \mathbb{Q}$ and $a + b + c = 0$.Since $a + b + c = 0$,we have $b = -(a + c)$.T

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Rational

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(A) Given the quadratic equation $ax^2 + bx + c = 0$ where $a, b, c \in \mathbb{Q}$ and $a + b + c = 0$.Since $a + b + c = 0$,we have $b = -(a + c)$.The discriminant $D$ is given by $D = b^2 - 4ac$.Substituting $b = -(a + c)$,we get $D = (-(a + c))^2 - 4ac = (a + c)^2 - 4ac$.Expanding this,$D = a^2 + 2ac + c^2 - 4ac = a^2 - 2ac + c^2 = (a - c)^2$.Since $a, c \in \mathbb{Q}$,$(a - c)^2$ is a perfect square of a rational number.For a quadratic equation with rational coefficients,if the discriminant $D$ is a perfect square of a rational number,the roots are rational.Therefore,both roots are rational.

List-$I$	List-$II$
$(i)$ $\alpha = \beta$	$(A)$ $(ac^2)^{1/3} + (a^2c)^{1/3} + b = 0$
$(ii)$ $\alpha = 2\beta$	$(B)$ $2b^2 = 9ac$
$(iii)$ $\alpha = 3\beta$	$(C)$ $b^2 = 6ac$
$(iv)$ $\alpha = \beta^2$	$(D)$ $3b^2 = 16ac$
	$(E)$ $b^2 = 4ac$
	$(F)$ $(ac^2)^{1/3} + (a^2c)^{1/3} = b$

If $a + b + c = 0$,$a \ne 0$,and $a, b, c \in \mathbb{Q}$,then both the roots of the equation $ax^2 + bx + c = 0$ are

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