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Solving the equation

  1. Jul 4, 2013 #1
    1. The problem statement, all variables and given/known data
    Find all real numbers ##x## such that,
    [tex]\left(x-\frac{1}{x}\right)^{1/2}+\left(1-\frac{1}{x}\right)^{1/2}=x[/tex]


    2. Relevant equations



    3. The attempt at a solution
    Moving the second term on LHS to the right and squaring both the sides,
    [tex]x-\frac{1}{x}=x^2+1-\frac{1}{x}-2x\sqrt{1-\frac{1}{x}}[/tex]
    Rearranging,
    [tex]x^2-x+1=2x\sqrt{1-\frac{1}{x}}[/tex]
    Squaring again leads to terms containing 4th and 3rd power of x and I don't find a way to proceed further. Here's the expression I get after squaring and rearranging,
    [tex]x^4+2x^3-x^2-2x+1=0[/tex]

    Any help is appreciated. Thanks!
     
  2. jcsd
  3. Jul 4, 2013 #2

    LCKurtz

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    ##x^4+2x^3-x^2-2x+1=(x^2+x-1)^2##
     
  4. Jul 4, 2013 #3
    [tex]\left(x-\frac{1}{x}\right)^{1/2}+\left(1-\frac{1}{x}\right)^{1/2}=x[/tex]

    Multiply both sides by [itex]\left(x-\frac{1}{x}\right)^{1/2}-\left(1-\frac{1}{x}\right)^{1/2}[/itex] to get

    [tex]\left(x-\frac{1}{x}\right)^{1/2}-\left(1-\frac{1}{x}\right)^{1/2}=1-\frac{1}{x}[/tex]
    Now, add the two equations to get:

    [tex]2\left(x-\frac{1}{x}\right)^{1/2}=\left(x-\frac{1}{x}\right)+1[/tex]

    Let [itex]y=\sqrt{\left(x-\frac{1}{x}\right)}[/itex]

    Then [tex]y^2-2y+1=0[/tex]
     
  5. Jul 5, 2013 #4
    That was too obvious to be missed. :tongue:

    Thank you! :smile:
    Nice one Chet, thanks a lot! :smile:
     
  6. Jul 5, 2013 #5


    Your final equation is incorrect. You need to redo the squaring and rearranging. The correct relationship is:
    [tex]x^4-2x^3-x^2+2x+1=0[/tex]
    This is the same as:
    [tex](x^2-x-1)^2=0[/tex]
     
  7. Jul 5, 2013 #6
    Sorry about the typo, I knew the correct expression. I redid the squaring and rearranging when I was following LCKurtz's hint. :)

    I shouldn't be posting questions in the midnight. :tongue:
     
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