Equation with high exponents

Question

I would appreciate any help with this problem: $ x^8+2x^7+2x^6+5x^5+3x^4+5x^3+2x^2+2x^1+1x^0=0 $

I know that when $x$ isn't zero $x^0=1$ so the equation could be re-written as $ x^8+2x^7+2x^6+5x^5+3x^4+5x^3+2x^2+2x+1=0 $. I am not sure what to do from here. I have tried using wolframalpha to get the solutions so I know real ones exist (2 of them, actually), but I have no idea how to get them.

score 5 · Answer 1 · answered May 26 '13 at 16:40

5

As the equation is Reciprocal Equation of the First type,

Divide either sides by $x^4,$ to get $$x^4+\frac1{x^4}+2\left(x^3+\frac1{x^3}\right)+2\left(x^2+\frac1{x^2}\right)+5\left(x+\frac1x\right)+3=0$$

Put $x+\frac1x=y$ to reduce the equation the degree $\frac82=4$

Can you take it form here?

answered May 26 '13 at 16:40

lab bhattacharjee

274,582

1

How many types are there, and what are the differences? - Perhaps this should be a separate question. – Mark Bennet May 26 '13 at 16:41
@MarkBennet, Please wait. I'm searching for an online link. – lab bhattacharjee May 26 '13 at 16:47
Alpha finds two real roots, one of which will yield two real roots for $x$, but they don't look simple. – Ross Millikan May 26 '13 at 16:49
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@MarkBennet, Chapter XI of http://archive.org/details/higheralgebra032813mbp and Article $568-570$ http://archive.org/details/higheralgebraseq00hall should explain it – lab bhattacharjee May 26 '13 at 18:13
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References like that have gone out of date, but contain fascinating material - thanks! – Mark Bennet May 26 '13 at 18:31

Equation with high exponents

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