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The square root of {y} is equal to the square root of {x²} which is also equal to 9.

 

\((x, \mbox{ } y) \in \mathbb{N}\)

\((x, \mbox{ } y) \in \mathbb{Z}\)

 Jan 7, 2022
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So you're actually solving \(\sqrt{y} = \sqrt{x^2} = 9\) .

 

Ignore \( \sqrt{x^2}\), then the equation would be \(\sqrt{y} = 9\) .

 

Now take the left and right sides to the power of 2, so \(y = 81\)

 

Now ignore \(\sqrt{y}\) and take \(\sqrt{x^2}\) in the equation again.

 

\(\sqrt{x^2} = 9\), it applies that \(\sqrt{x^2} = x\), then \(x = 9\) ,

 

so

 

x = 9 ,

y = 81,


x and y are positive integers.

 Jan 7, 2022

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