If `alpha, beta` are the roots of the equation `x^(2)-2x-a^(2)+1=0` and `gamma, delta` are the roots of the equation
`x^(2)-2(a+1)x+a(a-1)=0` such that `alpha, beta epsilonn (gamma, delta)` find the value of `a`.

Answer

`a epsilon(-1/4,1)`

Explanation

Since the given equation is
`x^(2)-2x-a^(2)+1=0`
`implies(x-1)^(2)=a^(2)`
`:.x-1!=a` or `x=1+-a`
`:.alpha=1+a` and `beta=1-a`

Let `f(x)=x^(2)-2(a+1)x+a(a-1)` thus the following conditiions hold good:
Consider the followign cases:
Case I `Dgt0`
`implies4(a+1)^(2)-4a(a-1)gt0`
`implies3a+1gt0`
`:.agt-1/3`
Case II` f(alpha)lt0`
`impliesf(1+a)lt0`
`implies(1+a)^(2)-2(1+a(1+a)+a(a-1)lt0`
`implies-(1+a)^(2)+a(a-1)lt0`
`implies-3a-1lt0`
`impliesagt-1/3`
Case II `f(s) =0`
`impliesf(1-a)lt0`
`implies(1-a)^(2)-2(a+1)(1-a)+a(a-1)lt0`
`implies(4a+1)(a-1)lt0`
`:.-1/4ltalt1`
Combining all cases we get
`a epsilon (-1/4,1)`