For how many real values of $x$ is $\sqrt{120-\sqrt{x}}$ an integer?
We know that for the square root of a number to be an integer, the number itself must be a perfect square. Right. So this means that $120 - \sqrt{x}$ must be a perfect square. We also know that perfect squares are non-negative. Thus $120 - \sqrt{x}$ is non-negative. Which means that $120 \ge \sqrt{x}$. That means that $120^2 \ge x$. But wait, we forgot to account for the fact that $\sqrt{x}$ is always non-negative. Right. So we also have to have $x \ge 0$. Now we have our two inequalities. Let's solve the first one for $x$. There are 11 of those. And the second one implies that we can look at all square numbers greater than 0. There are also 11 of those. So we have a total of 11 possible values of $x$. Notice that we did not have to find the values of $x$ themselves.

