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

MIRB16
Aug 19, 2017

edited by
Guest
Aug 19, 2017

#1**0 **

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