For what real values of `a` do the roots of the equation `x^2-2x-(a^2-1)=0` lie between the roots of the equation `x^2-2(a+1)x+a(a-1)=0.`

Explanation

`f_(1) (x) = x^(2) - 2x - (a^(2) - 1) = 0` (1)
`f_(2) (x) = x^(2) - 2(a + 1) x + a(a - 1) = 0 `(2)
From (1), we have
`x= (2pmsqrt(4 + 4(a^(2) - 1)))/(2) = 1 pm a`
Now, roots of (1) lie between roots of (2). Hence, graphs of functions `f_(1)(x) and f_(2)(x)` are as follwos :

From the graph, we have
`f_(2) (1-a)^(2) - 2(a+ 1)(1- a) + a(a - 1) lt 0`
`rArr (1 - a)^(2) - 2(a + 1) (1 - a) + a(a -1) lt 0`
or `(1 -a) [(1 -a) -2a -2 -a] lt0`
or `(1-a) (-4a-1) lt 0`
or ` (a - 1) (4a + 1) lt 0`
or -(1)/(4) lt a lt 1` (3)
and ` (1 + a)^(2) -2 (a +1)(a+1)+a(a-1) lt0`
or `-(a+ 2)^(2) + a (a - 1) lt 0`
or `-a^(2) - 2a - 1 + a^(2) -a lt 0`
or `3a + 1 gt 0`
or a `gt - (1)/(3)` (4)
From (3) and (4), the required values of a are `((-1)/(4), 1)`