Let alpha, beta be the roots of x^2 - x + p = | vedclass

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

Vedclass · Accepted Answer

$-2, -32$

Vedclass · Answer

(A) Let $r$ be the common ratio of the $G.P.$ $\alpha, \beta, \gamma, \delta$. Then $\beta = \alpha r, \gamma = \alpha r^2, \delta = \alpha r^3$.From the sum and product of roots:$\alpha + \beta = 1 \Rightarrow \alpha(1 + r) = 1$ $(i)$$\alpha \beta = p \Rightarrow \alpha^2 r = p$ $(ii)$$\gamma + \delta = 4 \Rightarrow \alpha r^2(1 + r) = 4$ $(iii)$$\gamma \delta = q \Rightarrow \alpha^2 r^5 = q$ $(iv)$Dividing $(iii)$ by $(i)$,we get $r^2 = 4$,so $r = \pm 2$.If $r = 2$,then $\alpha(1+2) = 1 \Rightarrow \alpha = 1/3$ (not an integer).If $r = -2$,then $\alpha(1-2) = 1 \Rightarrow \alpha = -1$.Substituting $r = -2$ and $\alpha = -1$ into $(ii)$ and $(iv)$:$p = (-1)^2(-2) = -2$$q = (-1)^2(-2)^5 = -32$Thus,$(p, q) = (-2, -32)$.

List-$I$	List-$II$
$(A)$ The minimum value of $2x^2 + 4x + 5$	$(I)$ $-1$
$(B)$ The maximum value of $\frac{x^2 + 4x + 1}{x^2 + x + 1}$	$(II)$ $1$
$(C)$ If $1 \leq \frac{3x^2 - 5x + 6}{x^2 + 1} \leq 2$, $\forall x \in [a, b]$ then $b =$	$(III)$ $2$
$(D)$ If $1 \leq \frac{3x^2 - 5x + 6}{x^2 + 1} \leq 2$, $\forall x \in [a, b]$ then $a =$	$(IV)$ $3$
	$(V)$ $4$

Let $\alpha, \beta$ be the roots of $x^2 - x + p = 0$ and $\gamma, \delta$ be the roots of $x^2 - 4x + q = 0$. If $\alpha, \beta, \gamma, \delta$ are in $G.P.$,then the integral values of $p, q$ are respectively:

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