+0

prob.

0
124
1

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

#1
+675
0

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