Let alpha and beta be the roots of x^{2}-3x+p | vedclass

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(C) Given the roots of $x^{2}-3x+p=0$ are $\alpha$ and $\beta$,and the roots of $x^{2}-6x+q=0$ are $\gamma$ and $\delta$.Since $\alpha, \beta, \gamma,

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$9:7$

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(C) Given the roots of $x^{2}-3x+p=0$ are $\alpha$ and $\beta$,and the roots of $x^{2}-6x+q=0$ are $\gamma$ and $\delta$.Since $\alpha, \beta, \gamma, \delta$ are in geometric progression,let them be $a, ar, ar^{2}, ar^{3}$.From the first equation,$\alpha+\beta = a+ar = 3$ and $\alpha\beta = a^{2}r = p$.From the second equation,$\gamma+\delta = ar^{2}+ar^{3} = 6$ and $\gamma\delta = a^{2}r^{5} = q$.Dividing the sum of roots of the second equation by the sum of roots of the first equation: $\frac{ar^{2}(1+r)}{a(1+r)} = \frac{6}{3} \implies r^{2} = 2$.Now,calculate $p$ and $q$ in terms of $a$ and $r$: $p = a^{2}r$ and $q = a^{2}r^{5} = a^{2}r(r^{2})^{2} = p(2)^{2} = 4p$.The ratio $\frac{2q+p}{2q-p} = \frac{2(4p)+p}{2(4p)-p} = \frac{8p+p}{8p-p} = \frac{9p}{7p} = \frac{9}{7}$.

Let $\alpha$ and $\beta$ be the roots of $x^{2}-3x+p=0$ and $\gamma$ and $\delta$ be the roots of $x^{2}-6x+q=0$. If $\alpha, \beta, \gamma, \delta$ form a geometric progression,then the ratio $(2q+p):(2q-p)$ is:

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