+287 Note that $\sqrt{x^2}$ is equivalent to $|x|$ so the question asks how many real $x$ makes $\sqrt{1200-|x|}$ an integer. Each such $x$ satisfies $1200 - |x| \ge 0$, and $1200-|x|$ must be a square number. Since $34^2 = 1156$ and $35^2 = 1225$, the possible values of $1200-|x|$ are $0^2, 1^2, 2^2, \ldots, 34^2$. Each of these square numbers gives exactly 2 values for $x$, one positive and one negative. E.g.
\[
0=1200-|1200|=1200-|-1200|
\]
and
\[
34^2=1200-|44|=1200-|-44|.
\]
So the answer is $35*2=70$.
+287 Note that $\sqrt{x^2}$ is equivalent to $|x|$ so the question asks how many real $x$ makes $\sqrt{1200-|x|}$ an integer. Each such $x$ satisfies $1200 - |x| \ge 0$, and $1200-|x|$ must be a square number. Since $34^2 = 1156$ and $35^2 = 1225$, the possible values of $1200-|x|$ are $0^2, 1^2, 2^2, \ldots, 34^2$. Each of these square numbers gives exactly 2 values for $x$, one positive and one negative. E.g.
\[
0=1200-|1200|=1200-|-1200|
\]
and
\[
34^2=1200-|44|=1200-|-44|.
\]
So the answer is $35*2=70$.
