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x +1/x - 1)^2 If x in the equation above does not equal one

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Manager
Joined: 12 Feb 2006
Posts: 115
x +1/x - 1)^2 If x in the equation above does not equal one [#permalink]

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02 Mar 2007, 03:17
This topic is locked. If you want to discuss this question please re-post it in the respective forum.

(x +1/x - 1)^2

If x in the equation above does not equal one or zero and is replaced with - 1/x, the result is which of the following?

(x +1/x - 1)^2
(x - 1/x + 1)^2
(x^2 + 1/1 - x^2)
(x^2 - 1/1 + x^2)
- (x - 1/x + 1)^2

I hope everything is clear. If not just let me know.

Last edited by bz9 on 02 Mar 2007, 23:14, edited 1 time in total.
VP
Joined: 29 Apr 2003
Posts: 1403

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02 Mar 2007, 03:23
A.

Are you sure it is replaced with 1/X and not -1/X?
Manager
Joined: 12 Feb 2006
Posts: 115

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02 Mar 2007, 23:15
Sorry, edited for clarity.

Can you guys work the problem?
Senior Manager
Joined: 18 Feb 2007
Posts: 253

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02 Mar 2007, 23:46
bz9 wrote:
Sorry, edited for clarity.

Can you guys work the problem?

is everything typed properly now??
Senior Manager
Joined: 29 Jan 2007
Posts: 441
Location: Earth

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03 Mar 2007, 10:19
if new values is (-1/x)

then I got B)

[(x-1)/(x+1)]^2
Manager
Joined: 09 Jan 2007
Posts: 240

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03 Mar 2007, 19:26
after replacement...
[{(-1/x)+1} / {(-1/x)-1}]^2
=> [{(-1+x)/x}/{(-1-x)/x}]^2
=> [(x-1)/(-x-1)]^2
=> {(-1)(x-1)/(x+1)}^2
=> {(x-1)/(x+1)}^2

Senior Manager
Joined: 18 Feb 2007
Posts: 253

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03 Mar 2007, 20:33
bz9: you have typed "(x +1/x - 1)^2 "
did you really mean to type: "((x + 1)/(x-1))^2" ?

the two are definitely not the same
Manager
Joined: 12 Feb 2006
Posts: 115

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03 Mar 2007, 22:08
Sorry,
I left this problem at work. I'll paste the entire question tomorrow.

My apologies
VP
Joined: 29 Apr 2003
Posts: 1403

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04 Mar 2007, 01:15
Agreed B with the new data
04 Mar 2007, 01:15
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