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For how many real values of x is an\(\sqrt{63-\sqrt{x}}\) integer?

MIRB16  Aug 19, 2017
edited by Guest  Aug 19, 2017
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For how many real values of x is does this expression have an integer value?

\(\sqrt{63-\sqrt{x}}\)

 

If x is an integer then the answer can be obtained by trial and error.

How many squated numbers are less then 63.

1,4,9,16,25,36,49,

63-62=1  x=62^2

63-59=4   x=59^2

63-...

..

63-14=49    x=14^2

 

So that is exactly 7 integer values of x will give the expression an integer value.

I do not think there are any other real values of x  that will work but I certainly have not proved that. ???

 

I'd be interested if someone expands on this. :)

Melody  Aug 20, 2017
edited by Melody  Aug 20, 2017

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