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find all roots of x^6-13^4+39x^2-27=0
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x^6-13^4+39x^2-27=0 ^ x there?
factor again if you missing the x there (x-1)*(x+3)*(x-3)*(x+1)*(x^2-3)
[x = 1], [x = -3], [x = 3], [x = -1], [x = sqrt(3)], [x = -sqrt(3)]
no their is an x right after the 13^4
you missing the x Jean :)
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yes I know, it is x^6-13^4x+39x^2-27=0
(x-1)*(x+3)*(x-3)*(x+1)*(x^2-3) =0
So would it be 1,-3,3,-1.
yes and sqrt(3) -sqrt(3)
YOU FORGET (x^2-3) =0 add 3 x^2=3 take sqrt x= plus or minute sqrt(3) -sqrt(3) sqrt(3)
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plus or minus*
Um so the answer is 1,sqrt(3), sqrt(3),-1, and x=plus or minus 3. I am a little confused.
1,-3,3,-1. are roots but also -sqrt(3) and sqrt(3)
okayyy. I get it now.
Good job Marge :D
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