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(A) Given that $\alpha, \beta$ are roots of $ax^2 + 2bx + c = 0$,so $\alpha + \beta = -\frac{2b}{a}$ and $\alpha\beta = \frac{c}{a}$.Given that $\gamm

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$q^2ac = b^2pr$

Vedclass · Answer

(A) Given that $\alpha, \beta$ are roots of $ax^2 + 2bx + c = 0$,so $\alpha + \beta = -\frac{2b}{a}$ and $\alpha\beta = \frac{c}{a}$.Given that $\gamma, \delta$ are roots of $px^2 + 2qx + r = 0$,so $\gamma + \delta = -\frac{2q}{p}$ and $\gamma\delta = \frac{r}{p}$.Since $\alpha, \beta, \gamma, \delta$ are in $G.P.$,let the common ratio be $k$. Then $\beta = \alpha k, \gamma = \alpha k^2, \delta = \alpha k^3$.From the roots,$\frac{\alpha\beta}{\gamma\delta} = \frac{c/a}{r/p} = \frac{cp}{ar}$.Substituting the $G.P.$ terms: $\frac{\alpha(\alpha k)}{\alpha k^2(\alpha k^3)} = \frac{\alpha^2 k}{\alpha^2 k^5} = \frac{1}{k^4} = \frac{cp}{ar}$.Also,$\frac{\alpha + \beta}{\gamma + \delta} = \frac{\alpha(1+k)}{\alpha k^2(1+k)} = \frac{1}{k^2}$.Substituting the sum of roots: $\frac{-2b/a}{-2q/p} = \frac{bp}{aq} = \frac{1}{k^2}$.Squaring this gives $\frac{b^2p^2}{a^2q^2} = \frac{1}{k^4}$.Equating the two expressions for $\frac{1}{k^4}$: $\frac{b^2p^2}{a^2q^2} = \frac{cp}{ar}$.Simplifying: $b^2p^2ar = a^2q^2cp \Rightarrow b^2pr = q^2ac$.

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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