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(Canada) Solve the equation

  • Thread starter Dinheiro
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  • #1
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Homework Statement


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]

Homework Equations


Algebric equations

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.
 

Answers and Replies

  • #2
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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.
 
  • #3
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Did you try working through the fractions? Then continue simplifying that way.
 
  • #4
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Did you try working through the fractions? Then continue simplifying that way.
Yes :/
 
  • #5
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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.
I couldn't get the solution from this
 
  • #6
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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?
etc.
 
  • #7
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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
 
  • #8
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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.
 
  • #9
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Yeah!! Thanks, guys, I got the answer: take y = x-1/x and have faith
 
Last edited:
  • #10
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Homework Statement


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]

Homework Equations


Algebric equations

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.
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]
Or,
[tex]\sqrt{(x+1)}-\sqrt{1-1/x} = 1[/tex]
Then, square both sides, and you obtain:

[tex]\left(x-\frac{1}{x}\right)-2\sqrt{\left(x-\frac{1}{x}\right)}+1=0[/tex]

This is a perfect square.
Chet
 
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  • #11
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awesome answer though, Chestermiller! I really appreciated this one xD
 

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