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Solving the equation

  • Thread starter Saitama
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  • #1
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Homework Statement


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]


Homework Equations





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!
 

Answers and Replies

  • #2
LCKurtz
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Homework Statement


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]


Homework Equations





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!
##x^4+2x^3-x^2-2x+1=(x^2+x-1)^2##
 
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  • #3
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[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]
 
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  • #4
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##x^4+2x^3-x^2-2x+1=(x^2+x-1)^2##
That was too obvious to be missed. :tongue:

Thank you! :smile:
[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]
Nice one Chet, thanks a lot! :smile:
 
  • #5
20,206
4,248
1.
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!


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]
 
  • #6
3,812
92
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]
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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