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Homework Help: (Canada) Solve the equation

  1. Jul 25, 2014 #1
    1. The problem statement, all variables and given/known data
    Find all real numbers [itex]x = \sqrt[2]{x-1/x} + \sqrt[2]{1-1/x}[/itex]

    * Note that [itex]\sqrt[2]{x-1/x}[/itex] is different from [itex]\sqrt[2]{1-1/x}[/itex]

    2. Relevant equations
    Algebric equations

    3. The attempt at a solution
    Squaring the equation will not really work, so I tried to substitute x for y = 1 - 1/x and, although I couldn't get the answer, it feels to be a good way: solve by substitution of variables.
  2. jcsd
  3. Jul 25, 2014 #2
    First note that x=0 is not a possible solution, since this equation involves terms of the form 1/x. Furthermore, when x≠0, we may multiply both sides of the equation by 1/x. Remember that [itex]a\sqrt{b} = \sqrt{a^2\cdot b}[/itex] when a ≥ 0.
  4. Jul 25, 2014 #3
    Did you try working through the fractions? Then continue simplifying that way.
  5. Jul 25, 2014 #4
    Yes :/
  6. Jul 25, 2014 #5
    I couldn't get the solution from this
  7. Jul 25, 2014 #6
    Also, what are you doing in terms of algebra? This will help us understand the context of the problem more.

    Are you allowed to use graphing calculators?
    Have you done Newton's Method?
  8. Jul 25, 2014 #7
    It is a mathematical olympiad problem and it isn't allowed anything, except for pencils and papers xD
    I am practicing with my cousin's book of polynomials and it says the answer is
    [itex]x = (1+ \sqrt{5})/2[/itex]
    Thanks for the help and patience, guys
  9. Jul 25, 2014 #8
    Okay, then first do the fractions under each radical. This will get you a sixth degree polynomial. You will then have to factor. My concern is what methods you know for factoring those things.
  10. Jul 25, 2014 #9
    Yeah!! Thanks, guys, I got the answer: take y = x-1/x and have faith
    Last edited: Jul 25, 2014
  11. Jul 26, 2014 #10
    Now that the problem has been solved, I'd like to provide an alternate solution. If I factor out [itex] \sqrt{1-1/x}[/itex] from the two terms on the RHS of the equation, I get:

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

    Next, multiplying numerator and denominator of the RHS by [itex]\left(\sqrt{(x+1)}-1\right)[/itex], I get:
    [tex]x = \sqrt{1-1/x}\frac{x}{\left(\sqrt{(x+1)}-1\right)}[/tex]
    Canceling the x from both sides of the equation then gives:
    [tex]\sqrt{(x+1)}-1 = \sqrt{1-1/x}[/tex]
    [tex]\sqrt{(x+1)}-\sqrt{1-1/x} = 1[/tex]
    Then, square both sides, and you obtain:


    This is a perfect square.
  12. Jul 26, 2014 #11
    awesome answer though, Chestermiller! I really appreciated this one xD
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