If alpha and beta are roots of ax^2 + 2bx + c | vedclass

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(C) Given the quadratic equation $ax^2 + 2bx + c = 0$.From the relation between roots and coefficients,we have $\alpha + \beta = -\frac{2b}{a}$ and $\

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$-\frac{2b}{\sqrt{ac}}$

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(C) Given the quadratic equation $ax^2 + 2bx + c = 0$.From the relation between roots and coefficients,we have $\alpha + \beta = -\frac{2b}{a}$ and $\alpha\beta = \frac{c}{a}$.Now,consider the expression $\sqrt{\frac{\alpha}{\beta}} + \sqrt{\frac{\beta}{\alpha}}$.Taking the common denominator,we get $\frac{\sqrt{\alpha} \cdot \sqrt{\alpha} + \sqrt{\beta} \cdot \sqrt{\beta}}{\sqrt{\alpha\beta}} = \frac{\alpha + \beta}{\sqrt{\alpha\beta}}$.Substituting the values of $\alpha + \beta$ and $\alpha\beta$,we get $\frac{-\frac{2b}{a}}{\sqrt{\frac{c}{a}}} = \frac{-\frac{2b}{a}}{\frac{\sqrt{c}}{\sqrt{a}}} = -\frac{2b}{a} \cdot \frac{\sqrt{a}}{\sqrt{c}} = -\frac{2b}{\sqrt{a} \cdot \sqrt{a}} \cdot \frac{\sqrt{a}}{\sqrt{c}} = -\frac{2b}{\sqrt{ac}}$.

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 $\alpha$ and $\beta$ are roots of $ax^2 + 2bx + c = 0$,then $\sqrt{\frac{\alpha}{\beta}} + \sqrt{\frac{\beta}{\alpha}}$ is equal to

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