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How to find the sum of the absolute values for the roots of this equation:


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up vote 7 down vote accepted

Notice that $$\begin{align}x^4-4x^3-4x^2+16x-8 &= (x-1)^4 - 10(x-1)^2 + 1 \\ &= ((x-1)^2-5)^2-24 \end{align}$$ so you can actually calculate the roots explicitly and sum their absolute values.

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Interestingly enough, some denesting must be happening, as the answer is $2\left(1+\sqrt{5+2\sqrt{6}}\right) = 2(1+\sqrt{2}+\sqrt{3})$. So maybe more subtle solution awaits its discovery. – Sasha Sep 21 '12 at 14:16
Do you have a neat trick to identify a shift that will work, or did you just try out simple cases to see if it happened to be one of them? – rschwieb Sep 21 '12 at 14:17
@rschwieb: In general, if $p(x)=x^n+ax^{n-1}+ \text{other stuff}$ is a degree-$n$ polynomial then, by the binomial theorem, the substitution $u=x+\frac{a}{n}$ clears the $(n-1)^{\text{th}}$ power (a neat trick I picked up when learning Galois theory). Here, all the multiples of $4$ enticed me into giving it a try, and it happened to work. – Clive Newstead Sep 21 '12 at 14:20
@CliveN. Interesting... thanks! When I looked at the coefficients, all I saw was that Eisenstein's criterion applied :) – rschwieb Sep 21 '12 at 14:36
@rschwieb: Are you sure? $2$ is the only prime dividing $8$, but $2^2$ also divides $8$ and so Eisenstein doesn't apply (at least, not with $p=2$). – Clive Newstead Sep 21 '12 at 14:40

Hint $\ $ Suppose that $\rm\ g(x\!+\!1)\, =\, f(x^2),\:$ and that $\rm\:f(x)\:$ has roots $\rm\:0< s < 1 < r.\:$

Then $\rm\:g\:$ has roots $\rm\:1\!-\!\sqrt{r}\, <\, 0\, <\, 1+\sqrt{r},\: 1\pm\sqrt{s},\:$ with absolute sum $\rm\ 2 + 2\,\sqrt{r}.$

In your case $\rm\:f(x) =\, x^2 - 10\,x + 1\:$ has roots $\rm\: 0 < 5 -2 \sqrt{6} < 1 < 5 + 2\sqrt{6}\:$ therefore, by above, we deduce that $\rm\:g\:$ has absolute root sum $= 2 + 2\sqrt{5+2\sqrt{6}}\, =\, 2\,(1 + \sqrt{2} + \sqrt{3}).$

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