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X^2тАУ4X|=3

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stolyar
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X^2тАУ4X|=3 [#permalink] New post   07 Jun 2003, 00:20
|X^2тАУ4X|=3



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Re: X^2тАУ4X|=3 [#permalink] New post   07 Jun 2003, 00:25
3,1,2+(7)^(1/2),2-(7)^(1/2)
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Re: X^2тАУ4X|=3 [#permalink] New post   07 Jun 2003, 00:26
Expert Reply
:shock: :beatup
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Re: X^2тАУ4X|=3 [#permalink] New post   07 Jun 2003, 07:20
There are 4 solutions:
X= 1

X= 3

X= 2+sqrt(7)

X= 2-sqrt(7)
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Re: X^2тАУ4X|=3 [#permalink] New post   14 Mar 2006, 20:10
hmm...isnt it just 1 and -4??
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Re: X^2тАУ4X|=3 [#permalink] New post   14 Mar 2006, 22:54
Negative case:

-x^2+4x = 3
x^2-4x+3 = 0
(x-1)(x-3)=0
x = 1, x = 3

Positive case:
x^2-4x=3
x^2-4x-3 = 0
x = 2 +/- sqrt(7)
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Re: X^2тАУ4X|=3 [#permalink] New post   15 Mar 2006, 20:35
In the positive case:

x^2-4x=3
x^2-4x-3 = 0
x = 2 +/- sqrt(7)

how do you factor the expression to get sqrt(7) ??? Not sure what a quick method is :?
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Re: X^2тАУ4X|=3 [#permalink] New post   15 Mar 2006, 20:58
Use the method x = (-b +/- sqrt (b^b - 4ac) ) / 2a to solve for those imaginery roots of x^x - 4x -3.

This gives you x = 2 + sqrt7 and 2 - sqrt7

x^x - 4x + 3 can be solved by normal factorisation to (X-3)(x-1) = 0, giving x=3 and x=1
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Re: X^2тАУ4X|=3 [#permalink] New post   15 Mar 2006, 21:47
I have never seen a root problem on the ETS version of the GMAT that required us to use the quadratic formula, but then again I never scored over Q42 either.
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Re: X^2тАУ4X|=3 [#permalink] New post   16 Mar 2006, 10:12
elektraa wrote:
Use the method x = (-b +/- sqrt (b^b - 4ac) ) / 2a to solve for those imaginery roots of x^x - 4x -3.

This gives you x = 2 + sqrt7 and 2 - sqrt7

x^x - 4x + 3 can be solved by normal factorisation to (X-3)(x-1) = 0, giving x=3 and x=1


A correction to your formula ---> it is b^2, not b^b.

Thanks for the explanation!

But do problems like this really show up on the GMAT, where we have to use the quadratic formula??



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Re: X^2тАУ4X|=3 [#permalink]
16 Mar 2006, 10:12
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