finding the sum of the absolute values for the roots How to find the sum of the absolute values for the roots of this equation:
$$x^4-4x^3-4x^2+16x-8=0$$

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 $= 2 + 2\sqrt{5+2\sqrt{6}}\, =\, 2\,(1 + \sqrt{2} + \sqrt{3}).$
