Root inside the Root

We note that $\sqrt{a-\sqrt{a^2+x}} \ge 0$ for real $a\ge 0$ . This implies that $x \ge 0$ . But if $x > 0$ , then $a-\sqrt{a^2+x} < 0$ and $\sqrt{a-\sqrt{a^2+x}}$ is not defined. Therefore, it can only be $\boxed{1}$ solution, $x=0$ .

Clearly $x = 0$ is a solution by inspection. We consider the cases $x > 0$ and $x < 0$ :

• $x>0$ : Then $\sqrt{a^2 + x} > \sqrt{a^2} = a$ , so $a - \sqrt{a^2 + x} < 0$ , thus $\sqrt{a - \sqrt{a^2 + x}}$ cannot be real.

• $x<0$ : Then $\sqrt{a - \sqrt{a^2 + x}} < 0$ , which is impossible.

Therefore $x = 0$ is the only solution.