Given that `alpha`, `gamma` are roots of the equation `Ax^(2)-4x+1=0` and `beta`, `delta` are roots of the equation `Bx^(2)-6x+1=0`. If `alpha`, `beta`,`gamma` and `delta` are in `H.P.`, then

Options

A. `A=5`
B. `A=3`
C. `B=8`
D. `B=-8`

Answer

B

Explanation

`(b,c)` `alpha,beta,gamma,delta` are in `H.P.`
`implies(1)/(alpha)`, `(1)/(beta)`, `(1)/(gamma)`, `(1)/(delta)` are in `A.P.`
Let `d` be the common difference of the `A.P.`
Since `alpha`, `gamma` are roots of `Ax^(2)-4x+1=0`
`:.(alpha+gamma)/(alphagamma)=(4//A)/(1//A)=4`
or `(1)/(alpha)+(1)/(gamma)=4`
`implies(1)/(alpha)+(1)/(alpha)+2d=4` or `(1)/(alpha)+d=2`.......`(i)`
Also `beta`, `delta` are roots of `Bx^(2)-6x+1=0`
`:.(beta+delta)/(betadelta)=(1)/(beta)+(1)/(delta)=(6//B)/(1//B)=6` or `(1)/(alpha)+d+(1)/(alpha)+3d=6`
`implies(1)/(alpha)+2d=3`..........`(ii)`
Solving `(i)` and `(ii)`, we get `(1)/(alpha)=1` and `d=1`
`:. (1)/(alpha)=1`, `(1)/(beta)=2`, `(1)/(gamma)=3` and `(1)/(delta)=4`
since `(1)/(alphagamma)=AimpliesA=3`
Also, `(betagamma)=BimpliesB=8`