6
$\begingroup$

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

$x^4-4x^3-4x^2+16x-8=0$

2 Answers 2

8

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.

  • 0
    Another decomposition you can use to solve the problem would be $P($x$)=$x$^2(x-2)^2-8(x-1)^2$.2012-09-21
2

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