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I have to solve the following equation for $x$, for all values of $a$: $$x+\sqrt{a+\sqrt{x}}=a$$

Clearly $x\gt0$

For $a\lt0,x\in\phi$ as RHS$\lt0$ but LHS$\gt0$.

$a=0\Rightarrow x=0$

Now for $a\gt0$ $$a-x=\sqrt{a+\sqrt{x}} ,x\in(0,a)$$ $$(a-x)^2=a+\sqrt{x}.....(1)$$ Now plotting the graphs for $$y=(a-x)^2...a\gt0,x\in(0,a)$$ $$y=a+\sqrt{x}$$ For $a\in(0,1)$ the graphs don't intersect$\Rightarrow x\in\phi$. Also, $a=1\Rightarrow x=0$.

For $a\gt1$ the graphs intersect so $$a^2-a+x^2-2ax=\sqrt{x}...from (1)$$Squaring and rearranging $$x^4-4ax^3+6a^2x^2-2ax^2-4a^3x+4a^2x+a^4-2a^3+a^2=x$$ which I am not able to solve further.

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    Substitute $y:=a-x$ and then it just becomes https://math.stackexchange.com/questions/1756108/real-solution-of-the-equation-sqrta-sqrta-x-x-if-a0?rq=1 – Anand Sep 22 '20 at 09:15

1 Answers1

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Let $x=(y^2-a)^2$ with $y^2\ge a$ and $y\ge 0$ then

$$x+\sqrt{a+\sqrt{x}}=a \iff (y^2-a)^2+y=a \iff y^4-2ay^2+y-a+a^2=0$$

and since the $y^3$ term is equal to zero we can guess

$$y^4-2ay^2+y-a+a^2=(y^2+y+A)(y^2-y+B)=$$

$$=y^4+(A+B-1)y^2+(B-A)y+AB$$

which leads to $A+B-1=-2a$, $B-A=1$, $AB=a^2-a$ that is

$$y^4-2ay^2+y-a+a^2=( y^2 + y-a) ( y^2 - y + 1-a) = 0$$

Answer completed with a full solution here.

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