Find the set of values for $c\in \mathbb R$ that allows real solution for $\sqrt{x}=\sqrt{\sqrt{x}+c}$

Question

Given $\{x,c\}\subset \mathbb R$, $\sqrt{x}=\sqrt{\sqrt{x}+c}~~~~ (1)$.

Find: set of values for $c$ such that $(1)$ has solution in $\mathbb R$.

Question from the Brazilian Math Olympiad 2004. No solution provided.

I'm not sure whether I'm setting the right constraints to find the asked set. Hints and full solutions are appreciated.

My attempt: as an initial observation, the solution must consider at least 2 constraints: $(c_1)$ $x\ge 0;$ and, $(c_2)$ $\sqrt{x}\ge -c$, to avoid negative arguments in the square root.

By assuming ($c_1$) and ($c_2$) hold, and squaring both terms in (1) we get $$\sqrt{x}=\sqrt{\sqrt{x}+c}\Leftrightarrow x=\sqrt{x}+c\Leftrightarrow x-c=\sqrt{x}.$$

Now considering an addicional constraint $(c_3)$ $x-c\ge 0$, and squaring both terms, we get $$x-c=\sqrt{x}\Leftrightarrow x^2-2cx+c^2=x \Leftrightarrow x^2-(2c+1)x+c^2=0$$ To solve this last equation, notice that $$\triangle=(2c+1)^2-4c^2=4c+1$$ Therefore, another constraint, for real roots, is $\triangle\ge 0$ or $(c_4)$ $c\ge -1/4$. And the tentative solution for $x$, before checking constraints, will be given by $$x=\frac{2c+1\pm\sqrt{4c+1}}{2}.$$ Now the set of values for $c$ that leads to a real solution in $x$, will be the set resulting from the intersection of 4 conditions: $$\left\{ \begin{array}{l} (c_1)~~x\ge 0 \Leftrightarrow \frac{2c+1\pm\sqrt{4c+1}}{2}\ge 0\Leftrightarrow 2c+1\ge\pm\sqrt{4c+1}\\ (c_2)~~\sqrt{x}\ge -c \Leftrightarrow \sqrt{\frac{2c+1\pm\sqrt{4c+1}}{2}}\ge -c\\ (c_3)~~x\ge c\Leftrightarrow \frac{2c+1\pm\sqrt{4c+1}}{2}\ge c\Leftrightarrow 1 \ge \pm\sqrt{4c+1}\Leftrightarrow 1 \ge \sqrt{4c+1}\Leftrightarrow c \le 0\\ (c_4)~~c\ge -1/4\\ \end{array} \right. $$ Question: (a) is this last step the right set of conditions to be intersected to give the asked set for $c$? (b) how to solve for the intersection of the conditions?

Hints and full answers are appreciated. Maybe I'm just complicating something that is a lot easier.

Answer 1

There is no problem ignoring the external square roots, as both are positive values.

Now

$$(\sqrt x)^2-\sqrt x-c=0$$

has real roots only when

$$1+4c\ge0,$$ and these roots are

$$\sqrt x=\frac{1\pm\sqrt{1+4c}}{2},$$ one of which is necessarily not smaller than $\dfrac12$.

With

$$c\ge-\frac14$$

both $x$ and $\sqrt x+c$ are guaranteed to be non-negative.

Answer 2

Just a slightly different point of view, which may be useful in a different situation. Since it is easier to express $c$ in terms of $x$ rather than the other way around, you might make this argument:
Let $f:\mathbb{R}_{\geq0} \to \mathbb{R}$ be given by $f(x) = x - \sqrt{x}$. Then $$\exists x\in \mathbb{R} (x\geq 0 \land c = x - \sqrt{x} = f(x)) \Leftrightarrow c \in {\rm ran} (f)$$ (there is no need for any other conditions because the domain of $f$ is exactly $x \geq 0$). Now here the substitution $\sqrt{x} = t$ would make finding the range more convenient. I will do it by noting that $g:\sqrt{x}\mapsto x$, with domain and codomain $\mathbb{R}_{\geq0}$, is surjective and so ${\rm ran}(g) = {\rm dom}(f)$; not only can the two functions be composed, but also ${\rm ran} (f\circ g) = {\rm ran} (f)$.

So instead of studying the solutions of an equation for some (unspecified) $c$, you could examine the function $f\circ g: \mathbb{R}_{\geq0} \to \mathbb{R}$, given by $f(g(t)) = t^2 - t$. (All the talk about domains and ranges in the previous paragraph was to demonstrate that no additional constraints are needed.) It's a quadratic restricted to $[0, +\infty)$ whose vertex is easily found to be $(1/2, -1/4)$ so the range is $[-1/4, +\infty)$.

This isn't really any different from finding the roots (parametrically) and checking/requiring that at least one be non-negative (as Yves does). But if it had been something more complicated than a quadratic, it is usually easier to find the range than to solve the corresponding equation.

EDIT: Note that when solving equations with radicals on both sides, $\sqrt{f(x)} = \sqrt{g(x)}$, it is redundant to add both $f(x) \geq 0$ and $g(x) \geq 0$. For any solution of the squared equation, $f(x) = g(x)$ they are equivalent. In your approach, $(c_1) \Leftrightarrow (c_2)$. And later on, the second squaring produces $x = (x-c)^2$ which automatically guarantees $(c_1)$. So only $(c_3)$ and $(c_4)$ are left, and it is not hard to check that the latter implies the former, since it is enough for one of the roots to meet it (I think you had already figured that out).