I Sqrt of y^2=x^2: Why y=x or y=-x?

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The equation y^2 = x^2 leads to |y| = |x|, which implies that y can be either equal to x or -x. This is derived from the fact that the absolute value equation |y| = |x| encompasses both positive and negative scenarios for y in relation to x. The discussion clarifies that while y could be expressed as |x| or -|x|, the solutions y = x and y = -x are the only valid interpretations when considering the square root. Furthermore, the factorization of y^2 - x^2 = 0 reinforces this conclusion by showing that y must equal x or -x. Thus, the relationship between y and x is defined by their equality or opposite signs.
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Say we have that ##y^2 = x^2##. Then if we take the square root of both sides, it would seem that we have ##|y| = |x|##. Why does this imply that that ##y=x## or ##y=-x##, rather than implying that ##y=|x|## or ##y=- |x|##?
 
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Mr Davis 97 said:
Say we have that ##y^2 = x^2##. Then if we take the square root of both sides, it would seem that we have ##|y| = |x|##. Why does this imply that that ##y=x## or ##y=-x##, rather than implying that ##y=|x|## or ##y=- |x|##?
The two answers are equivalent.

We agree that a = |x| means that
a = x or
a = -x.

If you let a = |y| then
(1) |y| = x or
(2) |y| = -x

But (1) x = |y| means x = y or x = -y and (2) -x = |y| means -x = y or -x = -y (i.e. x = y). So, y = x or y = -x.

AM
 
Mr Davis 97 said:
Say we have that ##y^2 = x^2##. Then if we take the square root of both sides, it would seem that we have ##|y| = |x|##. Why does this imply that that ##y=x## or ##y=-x##, rather than implying that ##y=|x|## or ##y=- |x|##?
Why not just do this?
##y^2 = x^2 \Leftrightarrow y^2 - x^2 = 0 \Leftrightarrow (y - x)(y + x) = 0 \Leftrightarrow y = x \text{ or } y = -x##
 
Yet another approach:
If |x|=|y|, there are four cases:
x and y positive: then x=y
x positive, y negative: x=-y
x negative, y positive: x=-y
x and y negative: x=y
Combined, x=y or x=-y. In other words, the two variables are identical up to a possible difference in their sign.

I neglected the option x=y=0 here, but that fits to the answer as well.
 
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