If alpha, beta are the roots of the equation | vedclass

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

Question

(C) Let the increasing $G.P.$ be $k, kr, kr^2, kr^3$ where $r > 1$ and $k > 0$.Then $\alpha = k, \beta = kr, \gamma = kr^2, \delta = kr^3$.From the fi

Vedclass · Accepted Answer

$(2, 32)$

Vedclass · Answer

(C) Let the increasing $G.P.$ be $k, kr, kr^2, kr^3$ where $r > 1$ and $k > 0$.Then $\alpha = k, \beta = kr, \gamma = kr^2, \delta = kr^3$.From the first equation $x^2 - 3x + a = 0$,the sum of roots $\alpha + \beta = k(1 + r) = 3$ and product $\alpha \beta = k^2 r = a$.From the second equation $x^2 - 12x + b = 0$,the sum of roots $\gamma + \delta = kr^2(1 + r) = 12$ and product $\gamma \delta = k^2 r^5 = b$.Dividing the sum equations: $\frac{kr^2(1 + r)}{k(1 + r)} = \frac{12}{3} \implies r^2 = 4$. Since the $G.P.$ is increasing,$r = 2$.Substituting $r = 2$ into $k(1 + r) = 3$,we get $k(3) = 3 \implies k = 1$.Thus,the roots are $1, 2, 4, 8$.Then $a = \alpha \beta = 1 	imes 2 = 2$ and $b = \gamma \delta = 4 	imes 8 = 32$.Therefore,$(a, b) = (2, 32)$.

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

Similar Questions

If $p, q, r$ are in a geometric progression and $a, b, c$ are in another geometric progression,then $cp, bq, ar$ are in...

The third term of a geometric progression is equal to the square of the first term. If its second term is $8$,then its sixth term will be:

If the roots of the equation $8 x^3+6 p x^2+3 q x-27=0$ are in a geometric progression,then $q^2+9 p^2+6 p q+q/p=$

If $y = x + x^2 + x^3 + \dots \infty$,then $x = $

If five $G.M.s$ are inserted between $486$ and $2/3$,then the fourth $G.M.$ will be:

Vedclass Test Series

Exam Paper Generator

Online Exam Module