If the equation `x^2-3p x+2q=0a n dx^2-3a x+2b=0` have a common roots and the other roots of the second equation is the reciprocal of the other roots of the first, then `(2-2b)^2` . `36p a(q-b)^2` b. `18p a(q-b)^2` c. `36b q(p-a)^2` d. `18b q(p-a)^2`

Options

A. `36 pa (q - b)^(2)`
B. `18 pa (q - b)^(2)`
C. `36 bq (p - a)^(2)`
D. `18 bq (p - a)^(2)`

Answer

3

Explanation

According to the question
`alpha, beta` are roots of equation `x^(2) - 3px + 2q = 0`
and `alpha, 1/beta` are roots of equation `x^(2) - 3ax + 2b = 0`
Therefore, we have `alpha + beta = 3p, alpha beta = 2q`
and `alpha + 1/beta = 3a, alpha/beta = 2b`
`therefore (2q - 2b)^(2) = (alphabeta - alpha/beta)^(2) = alpha^(2)(beta- 1/beta)^(2)`
`= alpha/beta * betaalpha[(alpha + beta) - (alpha + 1/beta)]^(2)`
`= (2b) (2q) [3p - 3a]^(2)`
`= 36 bq (p - a)^(2)`