I assume you did something like this?
We want two solutions, so the discriminant \(\Delta = b^2 - 4ac > 0\) - so we now have:
The thing is by doing all this, we're actually also adding a bunch of solutions in that first step: where the line \(kx\) cuts the negative root \(-\sqrt{x}-1\). Squaring both sides is dangerous in this way (in the same way taking a particular root removes some solutions). If it helps you should definitely do it graph it. Intuitively you should be thinking about this already (especially when you add 'solutions' like this) ie. are there any values that don't work in my proposed solution?
There are also other ways to verify this. Subbing certain values of \(k\) into \(k^2x^2+ (2k-1)x +1 = 0\) eg. -1 will yield you the equation \(x^2 -3x + 1 = 0\). In this example, there are two solutions for \(x\), but you'll find that subbing them into \(kx\) and \(\sqrt{x}-1\) will give you three different points (one in common, one unique to each equation).