$'a'$ for which roots of one equation lie between roots of other equation. If the range of values of $'a'$ for which the roots of the equation $x^2-2x-a^2+1=0$ lie between the roots of the equation $x^2-2(a+1)x+a(a-1)=0$ is $(p,q)$, find the value of $(q+\frac{1}{p^2})$
Could someone give me slight hint as I can't understand how to initiate.

$x^2-2x-a^2+1=0$.  Quadratic equation yields $x = \frac{2\pm \sqrt{4 - 4(a^+ 1)}}{2}$
$x^2-2(a+1)x+a(a-1)=0$.  Q.E. yields $x = \frac{2(a+1) \pm \sqrt{(2(a+1))^2 - 4a(a-1)}}{2}$
We are told $\frac{2(a+1) - \sqrt{(2(a+1))^2 - 4a(a-1)}}{2}< \frac{2- \sqrt{4 - 4(a^+ 1)}}{2}$ and $\frac{2+ \sqrt{4 - 4(a^+ 1)}}{2} <\frac{2(a+1) + \sqrt{(2(a+1))^2 - 4a(a-1)}}{2}$
We work that out to figure that $p \le a \le q$.  Then we evaluate $p + 1/q^2$.
