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# if x+(1/x) = 5, what is x^2+(1/x^2)

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Joined: 05 Apr 2005
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if x+(1/x) = 5, what is x^2+(1/x^2) [#permalink]

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24 Oct 2005, 19:41
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if x+(1/x) = 5, what is x^2+(1/x^2)?

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Senior Manager
Joined: 30 Oct 2004
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24 Oct 2005, 19:49
x + (1/x) = 5

x^2 + (1/x^2) = (x + 1/x)^2 - 2 = 5^2-2 = 25-2 = 23
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-Vikram

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24 Oct 2005, 20:45
23

Given x + 1/x = 5; Square on both sides..

(x + 1/x)^2 = 5^2
x^2 + 1/x^2 + 2*x*1/x = 25
x^2 + 1/x^2 + 2 = 25
x^2 + 1/x^2 = 25 - 2
x^2 + 1/x^2 = 23

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Manager
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24 Oct 2005, 20:47
vikramm wrote:
x + (1/x) = 5

x^2 + (1/x^2) = (x + 1/x)^2 - 2 = 5^2-2 = 25-2 = 23

Vikram, I agree with your answer, but I'm not sure what steps you took to get there. It looks like you're taking x^2 + (1/x^2) and making that equal to (x + 1/x)^2 (although I'm not sure where the -2 is coming from).
But, (x+1/x)^2 really works out to x^2 + 2/x + 1/x^4.

Here's how I came up with the answer:

x + (1/x) = 5, so 1/x = 5-x
also, if you multiply both side of the first equation by x, you get:
x^2 + 1 = 5x
x^2 = 5x-1

Plug the value for 1/x into the second equation:

x^2 + (5-x)^2 = x^2 + 25 - 10x + x^2
=2x^2 - 10x + 25

Then, plug the value for x^2 = 5x-1 into this last equation:
2(5x-1) - 10x + 25
=10x -2 - 10x + 25
= -2 + 25
= 23

UPDATED
D'oh, I see the dumb mistake I made now. My method, while it works, is unnecessarily long. Thanks to both of you who provided the shorter explanation.

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24 Oct 2005, 20:47
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