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(A) Given the roots $\alpha, \beta$ of $ax^2 + 2bx + c = 0$,we have $\alpha + \beta = -\frac{2b}{a}$ and $\alpha\beta = \frac{c}{a}$.Given the roots $

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$q^2ac = b^2pr$

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(A) Given the roots $\alpha, \beta$ of $ax^2 + 2bx + c = 0$,we have $\alpha + \beta = -\frac{2b}{a}$ and $\alpha\beta = \frac{c}{a}$.Given the roots $\gamma, \delta$ of $px^2 + 2qx + r = 0$,we have $\gamma + \delta = -\frac{2q}{p}$ and $\gamma\delta = \frac{r}{p}$.Since $\alpha, \beta, \gamma, \delta$ are in $G.P.$,let the common ratio be $k$. Then $\beta = \alpha k$,$\gamma = \alpha k^2$,and $\delta = \alpha k^3$.From the roots of the first equation: $\alpha + \alpha k = -\frac{2b}{a} \Rightarrow \alpha(1+k) = -\frac{2b}{a}$ and $\alpha^2k = \frac{c}{a}$.From the roots of the second equation: $\alpha k^2 + \alpha k^3 = -\frac{2q}{p} \Rightarrow \alpha k^2(1+k) = -\frac{2q}{p}$ and $\alpha^2k^5 = \frac{r}{p}$.Dividing the sum equations: $\frac{\alpha k^2(1+k)}{\alpha(1+k)} = \frac{-2q/p}{-2b/a} \Rightarrow k^2 = \frac{aq}{bp}$.Dividing the product equations: $\frac{\alpha^2k^5}{\alpha^2k} = \frac{r/p}{c/a} \Rightarrow k^4 = \frac{ar}{pc}$.Since $k^4 = (k^2)^2$,we have $(\frac{aq}{bp})^2 = \frac{ar}{pc} \Rightarrow \frac{a^2q^2}{b^2p^2} = \frac{ar}{pc}$.Simplifying,$a^2q^2pc = ar b^2p^2$ $\Rightarrow aq^2c = rb^2p$ $\Rightarrow q^2ac = b^2pr$.

Let $\alpha, \beta$ be the roots of the equation $ax^2 + 2bx + c = 0$ and $\gamma, \delta$ be the roots of the equation $px^2 + 2qx + r = 0$. If $\alpha, \beta, \gamma, \delta$ are in $G.P.$,then:

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