Algebra

If the square root of $(63 - 3\sqrt{x})$ is an integer, then it has to be a perfect square.

Here are perfect squares under 63:

49 = 7^2 

36 = 6^2

25 = 5^2

16 = 4^2

9 = 3^2

4 = 2^2

1 = 1^2

0 = 0^2

Since there are 8 perfect squares under 63, then there are 8 real values of $x$ where the $\sqrt{(63 - 3\sqrt{x})}$ is an integer.