Freeing up the variables of an equation and assigning signs to them

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Homework Statement



sqrtY^2=sqrtX^2

Solving for Y we get:

Y= X and Y= -X

Homework Equations


The Attempt at a Solution



Since both sides of sqrtY^2=sqrtX^2 are equal I thought the equation would solve simply as Y= X. Turns out it has also the second part Y= -X.

My questions are since both sides of the equation sqrtY^2= sqrtX^2 are EQUAL

1. why don't we just get Y=X

or

2. why Y also doesn't take the negative sign just to keep the both sides of the orginal equation in balance?

Thanks.
 
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Possible solutions of the equation:

y = 2 and x = -2;

y = 2 and x = 2;

y = -2 and x = 2.
 
graphs said:

Homework Statement



sqrtY^2=sqrtX^2

Solving for Y we get:

Y= X and Y= -X

Homework Equations





The Attempt at a Solution



Since both sides of sqrtY^2=sqrtX^2 are equal I thought the equation would solve simply as Y= X. Turns out it has also the second part Y= -X.

My questions are since both sides of the equation sqrtY^2= sqrtX^2 are EQUAL

1. why don't we just get Y=X

or

2. why Y also doesn't take the negative sign just to keep the both sides of the orginal equation in balance?

Thanks.
[itex]\sqrt{y^2} = \sqrt{x^2} \Rightarrow y^2 = x^2[/itex]
[itex]\Rightarrow y^2 - x^2 = 0 \Rightarrow (y - x)(y + x) = 0[/itex]
The solutions of the last equation are y = x and y = -x.
 
Mark44 said:
[itex]\sqrt{y^2} = \sqrt{x^2} \Rightarrow y^2 = x^2[/itex]
[itex]\Rightarrow y^2 - x^2 = 0 \Rightarrow (y - x)(y + x) = 0[/itex]
The solutions of the last equation are y = x and y = -x.

Amazing!
 
Mark44 said:
Why is that amazing?

Well, because I didn't even think about factoring and finding the zeros. Simple and elegant.

X^2- Y^2=0 was given.

What I did was X^2=Y^2, then sqrtX^2=sqrtY^2 to "liberate" the variables Y and X. Then I got stuck with the signs the variable X took, what with X being both positive and negative.

So someone, in another forum, explained it with "Because Y = -X is also a solution. √(2²) = √((-2)²), for example" which was very helpful too. I just never thought about factoring.

Anyway, thank you, people!
 
If you started with x2 - y2 = 0, then the quickest approach is to factor the left side, and not messing around taking square roots.