First we have to solve the two equations simultaneously so that the zeros of that equation are the points of intersection.
Sub

into
I suspect it is more efficient to recognise this as a quadratic in

and solve it directly as:
 \pm \sqrt{(1-2k)^2 - 4(1)(k^2 - 1)}}{2(1)} \\ &= \cfrac{2k - 1 \pm \sqrt{1-4k + 4k^2 - 4k^2 + 4}}{2} \\ &= \cfrac{2k - 1 \pm \sqrt{5 - 4k}}{2} \end{align*})
Now, if there is a solution to this equation with

then we have complex roots and hence there aren't two double roots. If there are two roots with

then we have four roots and thus no double roots. So, we need there to be exactly one value of x^2 and it can't be zero (draw a sketch to see why). For there only to be a single value of

, we require:
Perhaps more elegantly, examining
x^2+(k^2-1)=0)
, we can see that it must factorise as
\left(x^2 - B\right))
. This further factorises, provided

, as
\left(x - \sqrt{A}\right)\left(x + \sqrt{B}\right)\left(x - \sqrt{B}\right))
, and seen that it can only give two double roots if

, and so note that

.