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

Name: Exam Paper Generator | JEE NEET Paper Generator | Vedclass
Brand: Vedclass
Rating: 5 (35 reviews)

Question

(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 $r = 2$,$\alpha(1+2) = 1 \Rightarrow \alpha = 1/3$ (not an integer). If $r = -2$,$\alpha(1-2) = 1 \Rightarrow -\alpha = 1 \Rightarrow \alpha = -1$.Substituting $\alpha = -1$ and $r = -2$ into $(ii)$ and $(iv)$:$p = (-1)^2(-2) = -2$.$q = (-1)^2(-2)^5 = -32$.Thus,$(p, q) = (-2, -32)$.

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:

Similar Questions

If $x=\frac{\sqrt{3}+1}{\sqrt{3}-1}$ and $y=\frac{\sqrt{3}-1}{\sqrt{3}+1}$,find the value of $x^{2}+y^{2}$.

If the roots of the equation $lx^2 + nx + n = 0$ are in the ratio $p:q$,then $\sqrt{\frac{p}{q}} + \sqrt{\frac{q}{p}} + \sqrt{\frac{n}{l}} = $

If $a, b, c \in \mathbb{Q}$,then the roots of the equation $(b + c - 2a)x^2 + (c + a - 2b)x + (a + b - 2c) = 0$ are

$A$ real root of the equation $\log_{4}\{\log_{2}(\sqrt{x+8} - \sqrt{x})\} = 0$ is

The roots of the equation ${x^2} + 2\sqrt{3}x + 3 = 0$ are

Vedclass Test Series

Exam Paper Generator

Online Exam Module