Results 1 to 6 of 6

Math Help - lxl = lyl

  1. #1
    Newbie
    Joined
    Nov 2010
    Posts
    19

    lxl = lyl

    I wonna solve the equation lxl = lyl by using:
    sqrt(x^2) = sqrt(y^2)
    =>
    x^2 = y^2
    Here the easiest to do is to use the "zero-rule"
    x^2 - y^2 = (x-y)(x+y) = 0
    Where it's obvious that x=-y and y=-x etc.
    However I don't understand why the same result isnt obtainable by not using this rule for solving the equation. What goes wrong when you do this:
    x^2 = y^2
    =>
    x = (+-)sqrt(y^2)
    Where you can't really get any further without comming back to the original statement. What is it that the zero-rules enables you in this case? Like when u factor out and use it instead of dividing by x, which if x=0 is not allowed algebraically.
    Follow Math Help Forum on Facebook and Google+

  2. #2
    MHF Contributor
    Prove It's Avatar
    Joined
    Aug 2008
    Posts
    11,319
    Thanks
    1238

    Re: lxl = lyl

    Quote Originally Posted by aaaa202 View Post
    I wonna solve the equation lxl = lyl by using:
    sqrt(x^2) = sqrt(y^2)
    =>
    x^2 = y^2
    Here the easiest to do is to use the "zero-rule"
    x^2 - y^2 = (x-y)(x+y) = 0
    Where it's obvious that x=-y and y=-x etc.
    However I don't understand why the same result isnt obtainable by not using this rule for solving the equation. What goes wrong when you do this:
    x^2 = y^2
    =>
    x = (+-)sqrt(y^2)
    Where you can't really get any further without comming back to the original statement. What is it that the zero-rules enables you in this case? Like when u factor out and use it instead of dividing by x, which if x=0 is not allowed algebraically.
    Why do you even need to bother using this method? It's clear that the size of x is the same as the size of y, which means y = x or y = -x...
    Follow Math Help Forum on Facebook and Google+

  3. #3
    Member
    Joined
    Jul 2011
    Posts
    105
    Thanks
    1

    Re: lxl = lyl

    the whole state of affairs can be described by the following chain: |x|=|y|\Longleftrightarrow\sqrt {x^2} = \sqrt {y^2}\Longleftrightarrow x^2=y^2\Longleftrightarrow (x+y)(x-y)=0\Longleftrightarrow x=-y\vee x=y
    Follow Math Help Forum on Facebook and Google+

  4. #4
    Newbie
    Joined
    Nov 2010
    Posts
    19

    Re: lxl = lyl

    yeh okay. I just don't get why the method of solving x or y wont get you a result..
    Follow Math Help Forum on Facebook and Google+

  5. #5
    MHF Contributor
    Prove It's Avatar
    Joined
    Aug 2008
    Posts
    11,319
    Thanks
    1238

    Re: lxl = lyl

    Quote Originally Posted by aaaa202 View Post
    yeh okay. I just don't get why the method of solving x or y wont get you a result..
    Well psolaki showed you that it can, I'm just saying that understanding the modulus function means you don't need to go through that rigmarole...
    Follow Math Help Forum on Facebook and Google+

  6. #6
    MHF Contributor

    Joined
    Apr 2005
    Posts
    15,313
    Thanks
    1291

    Re: lxl = lyl

    Quote Originally Posted by aaaa202 View Post
    I wonna solve the equation lxl = lyl by using:
    sqrt(x^2) = sqrt(y^2)
    =>
    x^2 = y^2
    Here the easiest to do is to use the "zero-rule"
    x^2 - y^2 = (x-y)(x+y) = 0
    Where it's obvious that x=-y and y=-x etc.
    However I don't understand why the same result isnt obtainable by not using this rule for solving the equation.
    What goes wrong when you do this:
    x^2 = y^2
    =>
    x = (+-)sqrt(y^2)
    Where you can't really get any further without comming back to the original statement.
    Yes, you can. That says [tex]x= \pm|y|/tex] and, since |y| is either y or -y, you have x= y or x= -(y), x= (-y) or x= -(-y)= y giving the correct solutions. x= y or x= -y.

    What is it that the zero-rules enables you in this case? Like when u factor out and use it instead of dividing by x, which if x=0 is not allowed algebraically.
    I have no idea what you are asking here.
    Follow Math Help Forum on Facebook and Google+

Search Tags


/mathhelpforum @mathhelpforum