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

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Question

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

Vedclass · Accepted Answer

$a = 2, b = 32$

Vedclass · Answer

(C) Let $r > 1$ be the common ratio of the $G.P.$ $\alpha, \beta, \gamma, \delta$.Then $\beta = r\alpha, \gamma = r^2\alpha,$ and $\delta = r^3\alpha$.From the properties of roots of quadratic equations:$\alpha + \beta = \alpha(1 + r) = 3$ ..... $(i)$$\alpha \beta = \alpha^2 r = a$ ..... $(ii)$$\gamma + \delta = \alpha r^2(1 + r) = 12$ ..... $(iii)$$\gamma \delta = \alpha^2 r^5 = b$ ..... $(iv)$Dividing equation $(iii)$ by $(i)$,we get $\frac{\alpha r^2(1 + r)}{\alpha(1 + r)} = \frac{12}{3}$,which simplifies to $r^2 = 4$. Since the $G.P.$ is increasing,$r = 2$.Substituting $r = 2$ into equation $(i)$,we get $\alpha(1 + 2) = 3$,so $\alpha = 1$.Now,calculate $a$ and $b$:$a = \alpha^2 r = (1)^2(2) = 2$.$b = \alpha^2 r^5 = (1)^2(2^5) = 32$.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 that order) form an increasing $G.P.$,then:

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