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

  1. Nov 8, 2011 #1
    1. The problem statement, all variables and given/known data

    sqrtY^2=sqrtX^2

    Solving for Y we get:

    Y= X and Y= -X

    2. Relevant equations



    3. 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.
     
  2. jcsd
  3. Nov 8, 2011 #2
    Possible solutions of the equation:

    y = 2 and x = -2;

    y = 2 and x = 2;

    y = -2 and x = 2.
     
  4. Nov 8, 2011 #3

    Mark44

    Staff: Mentor

    [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.
     
  5. Nov 8, 2011 #4
    Amazing!!!
     
  6. Nov 8, 2011 #5

    Mark44

    Staff: Mentor

    Why is that amazing?
     
  7. Nov 8, 2011 #6
    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!
     
  8. Nov 8, 2011 #7

    Mark44

    Staff: Mentor

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