If alpha, beta are the roots of x^2 - 3x + a | vedclass

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(C) Let $r > 1$ be the common ratio of the $G.P.$ $\alpha, \beta, \gamma, \delta$.Then $\beta = r\alpha, \gamma = r^2\alpha, \delta = r^3\alpha$.From

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$a = 2, b = 32$

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(C) Let $r > 1$ be the common ratio of the $G.P.$ $\alpha, \beta, \gamma, \delta$.Then $\beta = r\alpha, \gamma = r^2\alpha, \delta = r^3\alpha$.From the sum of roots for the first equation: $\alpha + \beta = \alpha(1 + r) = 3$ $(i)$.From the product of roots for the first equation: $\alpha\beta = \alpha^2r = a$ $(ii)$.From the sum of roots for the second equation: $\gamma + \delta = \alpha r^2(1 + r) = 12$ $(iii)$.From the product of roots for the second equation: $\gamma\delta = \alpha^2r^5 = b$ $(iv)$.Dividing $(iii)$ by $(i)$: $\frac{\alpha r^2(1 + r)}{\alpha(1 + r)} = \frac{12}{3} \Rightarrow r^2 = 4$. Since the $G.P.$ is increasing,$r = 2$.Substituting $r = 2$ into $(i)$: $\alpha(1 + 2) = 3$ $\Rightarrow 3\alpha = 3$ $\Rightarrow \alpha = 1$.Now,$a = \alpha^2r = (1)^2(2) = 2$.And $b = \alpha^2r^5 = (1)^2(2^5) = 32$.Thus,$a = 2$ and $b = 32$.

If $\alpha, \beta$ are the roots of $x^2 - 3x + a = 0$ and $\gamma, \delta$ are the roots of $x^2 - 12x + b = 0$,and the numbers $\alpha, \beta, \gamma, \delta$ (in order) form an increasing $G.P.$,then:

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