A radical equation with parameter

Condition. $\sqrt{x+a}=x$ requires $x\ge0$ and $x+a\ge0$. Squaring: $x+a=x^2$, i.e. $x^2-x-a=0$.

Roots. $x=\frac{1\pm\sqrt{1+4a}}{2}$, requiring $a\ge-1/4$.

Condition $x\ge0$. With $a\ge-1/4$, the root $x_+=\frac{1+\sqrt{1+4a}}{2}>0$. So $x_+$ is valid provided $x_+^2-x_+-a=0$ and $x_++a\ge0$. From $x^2=x+a$, $x+a=x^2\ge0$ automatically. So $x_+$ works.

The root $x_-=\frac{1-\sqrt{1+4a}}{2}$ is $\ge0$ iff $\sqrt{1+4a}\le1$, i.e. $a\le0$. And $a\ge-1/4$.

Conclusion. For $a\ge-1/4$, the equation has at least one solution (namely $x_+$). For $a\in[-1/4,0]$, it has two solutions ($x_+$ and $x_-$).