In the equatioin x^2 + 1/x^2=10/3, what values can x be : Quant Question Archive [LOCKED]
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# In the equatioin x^2 + 1/x^2=10/3, what values can x be

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In the equatioin x^2 + 1/x^2=10/3, what values can x be [#permalink]

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22 Feb 2006, 03:00
This topic is locked. If you want to discuss this question please re-post it in the respective forum.

In the equatioin x^2 + 1/x^2=10/3, what values can x be?
Director
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22 Feb 2006, 03:17
I don't know how to deal with the 1/(x^2):

x^2 + 1/x^2=10/3

(x + 1/x)^2 - 2= 10/3

(x + 1/x)^2 = 16/3

how to proceed?
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22 Feb 2006, 06:57
Multiply each side by x^2. You then have a quadratic equation to solve.
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22 Feb 2006, 11:58
I don't know how to deal with the 1/(x^2):

x^2 + 1/x^2=10/3

(x + 1/x)^2 - 2= 10/3

(x + 1/x)^2 = 16/3

how to proceed?

How come we're subtracting 2?
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22 Feb 2006, 12:32
petefroml wrote:
Multiply each side by x^2. You then have a quadratic equation to solve.

thanks

We add -2+2=0 to the equation so that we can simplify

x^2 +2+ 1/x^2 -2=10/3

(x + 1/x)^2 - 2= 10/3

Ok?
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22 Feb 2006, 13:07
I saw this question somewhere but a bit different and than it makes sense to calulate

x^2 + 1/x^2=2 what is x^4+1/x^4=?
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22 Feb 2006, 14:08
Here is how I would do it:
x^2 + 1/x^2=10/3
x^4 + 1 = 10/3*x^2
3x^4 - 10x^2 + 3 = 0

U=x^2:
3U^2 - 10U + 3 = 0
U = 3 or 1/3

x= sqr(3 ), -sqr(3 ), 1/sqr(3) or -1/sqr(3)
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22 Feb 2006, 14:11
x^2 + 1/x^2=10/3

x^4 + 1=10x^2/3

take x^2 as y then equation becomes

y^2 - 10y/3 + 1 = 0
(y-1/3) (y-3) = 0

so y = 1/3 or y = 3
so x = 1/SQRT(3) , -1/SQRT(3) or x = SQRT(3), -SQRT(3)
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Director
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23 Feb 2006, 04:40
oa is x= sqr(3 ), -sqr(3 ), 1/sqr(3) or -1/sqr(3)

great explanations
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23 Feb 2006, 18:47
i would solve it just like petefroml
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23 Feb 2006, 18:55
x^2 + 1/x^2=10/3

(x^4 + 1)/x^2 = 10/3
3x^4 + 3 = 10x^2
Replace x^2 with y

3y^2 -10y +3 = 0

(3y-1)(y-3) = 0

y = 1/3 or y = 3

x^2 = 1/3 or x^2 = 3
x = +/- sqrt(1/3), +/- sqrt(3)
23 Feb 2006, 18:55
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