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For how many real values of x is sqrt(63 - 3*sqrt(x)) an integer?

 Apr 27, 2021
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You cannot find the square root of a negative number 

So x is a positive square number.

 

and     \(0\le 63-3\sqrt{x} \le63\)

 

Now just test integer square values of x starting with x=0, then 1,4,9 etc

 Apr 28, 2021

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