prove that for $-1/4 < a < 0$, the equation $\sqrt{a + \sqrt{x+a}} = x$ has exactly 2 solutions one of which is $(1+\sqrt{4a+1})/2$. prove that for $-1/4 < a < 0$, the equation $\sqrt{a + \sqrt{x+a}} = x$ has exactly 2 solutions one of which is $(1+\sqrt{4a+1})/2$.
I have tried squaring both sides and obtianed a biquadratic equation. According to the question it should have 2 solutions. From the given equation it is evident that the solutions will be positive. After this can not figure out how to proceed. 
$\sqrt{a+\sqrt{x+a}} = x$
$\sqrt{a+x} = x^2 -a$
$x+a = (x^2-a)^2$
need help

We have $$\sqrt{a+x}= x^2-a\implies a^2-a(2x^2+1)+x^4-x=0$$
Solving this equation on $a$ we get, discriminat = $(2x+1)^2$ so $$ a= {2x^2+1\pm (2x+1)\over 2}$$


*

*case: If $a= x^2+x+1$ we get $$ x= {-1\pm \sqrt{4a-3}\over 2}$$
which does not count since $a< 0$.

*case: If $a= x^2-x$ we get $$ x= {1\pm \sqrt{4a+1}\over 2}$$

