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For how many real values of x is sqrt(1200 - sqrt(x^2)) an integer?

 Jun 14, 2021

Best Answer 

 #1
avatar+287 
+3

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 $35*2=70$.

 Jun 14, 2021
 #1
avatar+287 
+3
Best Answer

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 $35*2=70$.

Bginner Jun 14, 2021
 #2
avatar+121063 
0

Very nice, Bginner   !!!!

 

 

cool cool cool

CPhill  Jun 14, 2021

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