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(C) Given that $\alpha$ and $\beta$ are the roots of the equation $ax^2 + bx + c = 0$.From the relation between roots and coefficients,we have $\alpha

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$\frac{1}{\alpha}, \frac{1}{\beta}$

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(C) Given that $\alpha$ and $\beta$ are the roots of the equation $ax^2 + bx + c = 0$.From the relation between roots and coefficients,we have $\alpha + \beta = -\frac{b}{a}$ and $\alpha\beta = \frac{c}{a}$.Let the roots of the equation $cx^2 + bx + a = 0$ be $\alpha'$ and $\beta'$.Then,$\alpha' + \beta' = -\frac{b}{c}$ and $\alpha'\beta' = \frac{a}{c}$.We observe that $\alpha'\beta' = \frac{a}{c} = \frac{1}{c/a} = \frac{1}{\alpha\beta}$.Also,$\alpha' + \beta' = -\frac{b}{c} = \frac{-b/a}{c/a} = \frac{\alpha + \beta}{\alpha\beta} = \frac{1}{\beta} + \frac{1}{\alpha}$.Thus,the roots are $\frac{1}{\alpha}$ and $\frac{1}{\beta}$.

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 the roots of the equation $ax^2 + bx + c = 0$ are $\alpha$ and $\beta$,then the roots of the equation $cx^2 + bx + a = 0$ are

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