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# 6x^4 - 25x^3 + 12x^2 + 25x + 6 = 0 Find possible values of

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Manager
Joined: 28 Jun 2004
Posts: 90

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6x^4 - 25x^3 + 12x^2 + 25x + 6 = 0 Find possible values of [#permalink]

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25 Aug 2004, 15:47
This topic is locked. If you want to discuss this question please re-post it in the respective forum.

6x^4 - 25x^3 + 12x^2 + 25x + 6 = 0

Find possible values of x.

(Here x^n = x*x*....n times).

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Senior Manager
Joined: 05 Feb 2004
Posts: 290

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Location: USA

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26 Aug 2004, 07:24
x= 2, -1/2, 3, -1/3.....!!

6x^4 - 25x^3 + 12x^2 + 25x + 6 = 0
=>6x^4+12x^2+6-25x(x^2-1)=0
=>6(x^2+1)^2 -25x(x^2-1)=0
=>6{(x^2-1)^2 +4x^2} -25x(x^2-1)=0
Let (x^2-1)=a

=>6(a^2+4x^2)-25ax=0
=>6a^2-25ax+24x^2 =0
=>(3a-8x)(2a-3x)=0
=>a=8x/3 or 3x/2!!
Substitute a = x^2-1 and u have 2 quadratics and hence the 4 values which result upon solving the quadratics!!!

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Intern
Joined: 22 Mar 2004
Posts: 25

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Location: TX

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26 Aug 2004, 08:44
cbrf3,
you just showed that there is always an easy way to solve these kind of equations. i wonder if there are any other ways to solve them.

longhorn2020

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26 Aug 2004, 08:44
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