Question

Let `alphaa n dbeta` be the roots of `x^2-x+p=0a n dgammaa n ddelta` be the root of `x^2-4x+q=0.` If `alpha,beta,a n dgamma,delta` are in G.P., then the integral values of `pa n dq` , respectively, are `-2,-32` b. `-2,3` c. `-6,3` d. `-6,-32`
A. `-2, -32`
B. `-2, 3`
C. `-6, 3`
D. `-6, -32`

Answer 1

Correct Answer - A
Since `alpha, beta` are the roots of the equation `x^(2) - x + p = 0 and gamma, delta` are the roots of the equation `x^(2) - 4x + q = 0`. Therefore,
`alpha + beta = 1, alpha beta = p, gamma + delta =4, gamma delta = q`
Let r be the common ratio of the G.P. `alpha, beta, gamma, delta`. Then,
`beta = alpha r, gamma = alpha r^(2) and delta = alpha r^(3)`.
Now,
`alpha + beta = 1 rArr alpha + alpha r = 1`
`gamma + delta = 4 rArr alpha r^(2) + alpha r^(3) = 4`
`therefore" "(alpha r^(2) + alpha r^(3))/(alpha + alpha r) = 4 rArr r = +- 2`.
CASE I When r = 2
In this case, we have `alpha + beta = 1`
`rArr" "alpha + alpha r = 1 rArr 3 alpha = 1 rArr alpha = (1)/(3)" "[because r = 2]`
`therefore" "p = alpha beta = alpha^(2) r = (2)/(9)`
But, p is an integer. Therefore, r = 2 is not possible.
CASE II When r = - 2
In this case, we have `alpha + beta = 1`
`rArr" "alpha + alpha r = 1 rArr -alpha = 1 rArr alpha = -1" "[because r = -2]`
`therefore" "alpha + beta = 1 rArr beta = 2`
Now, `p = alpha beta rArr p = - 2`
and, `q = gamma delta = (alpha r^(2))(alpha r^(3)) = - 32`
Hence, `p = -2 and q = -32`.