prove that for $-1/4 < a < 0$, the equation $\sqrt{a + \sqrt{x+a}} = x$ has exactly 2 solutions one of which is $(1+\sqrt{4a+1})/2$.

Question

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

Answer 1

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 $$ a= {2x^2+1\pm (2x+1)\over 2}$$

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