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 $k > 0$ and $r > 1$.Thus,$\alpha = k, \beta = kr, \gamma = kr^2, \delta = kr^3$.From the fi

Vedclass · Accepted Answer

$(2, 32)$

Vedclass · Answer

(C) Let the increasing $G.P.$ be $k, kr, kr^2, kr^3$ where $k > 0$ and $r > 1$.Thus,$\alpha = k, \beta = kr, \gamma = kr^2, \delta = kr^3$.From the first equation $x^2 - 3x + a = 0$,the sum of roots $\alpha + \beta = k(1 + r) = 3$ and the product $\alpha \beta = k^2r = a$.From the second equation $x^2 - 12x + b = 0$,the sum of roots $\gamma + \delta = kr^2(1 + r) = 12$ and the product $\gamma \delta = k^2r^5 = b$.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 $k(3) = 3 \implies k = 1$.Now,$a = k^2r = (1)^2(2) = 2$.And $b = k^2r^5 = (1)^2(2^5) = 32$.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 $\alpha, \beta, \gamma, \delta$ form an increasing $G.P.$,then $(a, b) = $

Similar Questions

The sum of all two-digit positive numbers which, when divided by $7$, yield $2$ or $5$ as a remainder is:

Consider two positive numbers $a$ and $b$. If the arithmetic mean of $a$ and $b$ exceeds their geometric mean by $\frac{3}{2}$ and the geometric mean of $a$ and $b$ exceeds their harmonic mean by $\frac{6}{5}$,then the absolute value of $(a^2 - b^2)$ is equal to

Find the $25^{th}$ common term of the following $A.P.'s$
$S_1 = 1, 6, 11, .....$
$S_2 = 3, 7, 11, .....$

The mean of the series $a, a + nd, a + 2nd$ is

The product of $n$ positive numbers is unity. Their sum is

Vedclass Test Series

Exam Paper Generator

Online Exam Module