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

 Jan 25, 2022
 #1
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Seven

 Jan 25, 2022
 #2
avatar+834 
+3

If the square root of \((63 - 3\sqrt{x})\) is an integer, then it has to be a perfect square.

 

Here are perfect squares under 63:

 

49 = 7^2 

36 = 6^2

25 = 5^2

16 = 4^2

9 = 3^2

4 = 2^2

1 = 1^2

0 = 0^2

 

Since there are 8 perfect squares under 63, then there are 8 real values of \(x\) where the \(\sqrt{(63 - 3\sqrt{x})}\) is an integer.

 

 

smiley

 Jan 25, 2022

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