<s>[INST] Solving for $k$: $\sqrt{k-\sqrt{k+x}}-x = 0$ 
$$\sqrt{k-\sqrt{k+x}}-x = 0$$
Solve for $k$ in terms of $x$

I got all the way to
$$x^{4}-2kx^{2}-x+k^{2}-x^{2}$$
but could not factor afterwards. My teacher mentioned that there was grouping involved
Thanks Guys!
Edit 1 : The exact problem was solve for $x$ given that $$\sqrt{4-\sqrt{4+x}}-x = 0$$ with a hint of substitute 4 with k
 [/INST] As discovered $\sqrt{k - \sqrt{k+x}} = x$ leads to the equation $k^2 - (2 \, x^2 + 1) \, k + (x^4 - x) = 0$. Now,
\begin{align}
k &= \frac{1}{2} \, \left[(2 \, x^2 + 1) \pm \sqrt{ (2 \, x^2 + 1)^2 - 4 \, (x^4 - x) } \right] \\
&= \frac{1}{2} \, [(2 \, x^2 + 1) \pm (2 x + 1)] \\
&= \begin{cases}{ x^2 + x + 1 = \frac{1 - x^3}{1-x} \\ x(x-1) } \end{cases} 
\end{align} 
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