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# help algebra

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

Jun 14, 2021

#1
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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
+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 $35*2=70$.

Bginner Jun 14, 2021
#2
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Very nice, Bginner   !!!!   CPhill  Jun 14, 2021