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

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Manager
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6x^4 - 25x^3 + 12x^2 + 25x + 6 = 0 Find possible values of [#permalink]  25 Aug 2004, 14:47
6x^4 - 25x^3 + 12x^2 + 25x + 6 = 0

Find possible values of x.

(Here x^n = x*x*....n times).
Senior Manager
Joined: 05 Feb 2004
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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!!!
Intern
Joined: 22 Mar 2004
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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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