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

Options

A. `A=3`
B. `A=4`
C. `B=2`
D. `B=8`

Explanation

Since `alpha,beta, gamma` and `delta` are in HP hence `1/(alpha),1/(beta),1/(gamma)` and `1/(delta)` are in AP and they may be take as a `a-3d, a-d, a+d` and `a+3d`. Replaceing `x` by `1/x` we get the equation whose roots are `a-3d, a+d` is `x^(2)-4x+A=0` and equation whose roots are `a-d,a+3d` is `x^(2)-6x+B=0` then
`(a-3d)+(a+d)=4implies2(a-d)=4`
and `(a-d)+(a+3d)=6implies2(a+d)=6`
`:.a=5/2` and `d=1/2`
Now `A=(a-3d)(a+d)=(5/2-3/2)(5/2+1/2)=3`
and `B=(a-d)(a+3d)(5/2-1/2)(5/2+3/2)=8`