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 one root of the equation $ax^2 + bx + c = 0$ is $n$ times the other root,then:

Number of integral values of $a$ for which both roots of the quadratic equation $x^2 - (2a + 3)x + a^2 + 3a = 0$ lie in the interval $(0, 4)$ is:

If $(x + 1)$ is a factor of ${x^4} - (p - 3){x^3} - (3p - 5){x^2} + (2p - 7)x + 6$,then $p = $

If $\alpha$ and $\beta$ are the roots of the equation $x^2 - 2x + 3 = 0$,then the equation whose roots are $\frac{1}{\alpha^2}$ and $\frac{1}{\beta^2}$ is:

If $l, m, n$ are real and $l \ne m$,then the roots of the equation $(l - m)x^2 - 5(l + m)x - 2(l - m) = 0$ are

Vedclass Test Series

Exam Paper Generator

Online Exam Module