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# X^2тАУ4X|=3

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SVP
Joined: 03 Feb 2003
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07 Jun 2003, 00:20
This topic is locked. If you want to discuss this question please re-post it in the respective forum.

|X^2тАУ4X|=3

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Manager
Joined: 25 May 2003
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07 Jun 2003, 00:25
3,1,2+(7)^(1/2),2-(7)^(1/2)

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Founder
Joined: 04 Dec 2002
Posts: 15590

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Location: United States (WA)
GMAT 1: 750 Q49 V42

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07 Jun 2003, 00:26

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Intern
Joined: 05 Apr 2003
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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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Anh

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Intern
Joined: 18 Feb 2005
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14 Mar 2006, 20:10
hmm...isnt it just 1 and -4??

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GMAT Club Legend
Joined: 07 Jul 2004
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Location: Singapore

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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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Manager
Joined: 30 Jan 2006
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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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Intern
Joined: 15 Mar 2006
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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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Current Student
Joined: 29 Jan 2005
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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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Manager
Joined: 30 Jan 2006
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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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16 Mar 2006, 10:12
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