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

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(A) Given $\alpha + \beta = -\frac{b}{a}$,$\alpha \beta = \frac{c}{a}$ and $\gamma + \delta = -\frac{q}{p}$,$\gamma \delta = \frac{r}{p}$.Since $\alph

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$\frac{a^2}{p^2}$

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(A) Given $\alpha + \beta = -\frac{b}{a}$,$\alpha \beta = \frac{c}{a}$ and $\gamma + \delta = -\frac{q}{p}$,$\gamma \delta = \frac{r}{p}$.Since $\alpha, \beta, \gamma, \delta$ are in arithmetic progression,the common difference is constant.Thus,$\beta - \alpha = \delta - \gamma$.Squaring both sides,we get $(\beta - \alpha)^2 = (\delta - \gamma)^2$.Using the identity $(x - y)^2 = (x + y)^2 - 4xy$,we have $(\alpha + \beta)^2 - 4\alpha \beta = (\gamma + \delta)^2 - 4\gamma \delta$.Substituting the values,we get $\frac{b^2}{a^2} - \frac{4c}{a} = \frac{q^2}{p^2} - \frac{4r}{p}$.This simplifies to $\frac{b^2 - 4ac}{a^2} = \frac{q^2 - 4pr}{p^2}$.Since $D_1 = b^2 - 4ac$ and $D_2 = q^2 - 4pr$,we have $\frac{D_1}{a^2} = \frac{D_2}{p^2}$.Therefore,$\frac{D_1}{D_2} = \frac{a^2}{p^2}$.

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 $\alpha, \beta$ are the roots of $ax^2 + bx + c = 0$ and $\gamma, \delta$ are the roots of $px^2 + qx + r = 0$,and $D_1, D_2$ are their respective discriminants. If $\alpha, \beta, \gamma, \delta$ are in arithmetic progression,then $D_1 : D_2 = \dots$

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