MHB Find the sum of the real roots

[anemone]

Find the sum of the real roots for $2x^8-9x^7+20x^6-33x^5+46x^4-66x^3+80x^2-72x+32=0$.

 

[Fallen Angel]

Hi,

Spoiler

It is easy to see that $1,2,1+i,1-i$ are roots of the polynomial.

The quotient of this polynomial over the corresponding factors is $p(x)=2x^4+x^3+5x^2+2x+8$.

It is clear that $p(x)>0, \ \forall x\geq 0$.

If $x\in [-1,0)$ then $|x^3+2x|<8$ so $p(x)>0$.

If $x<-1$ then $2x^4>|x^3|$ and $5x^2>|2x|$ so $p(x)>0$.

Hence the sum is 3.

 

[kaliprasad]

Good solution by Fallen angel

here is mine

Spoiler

we have
$2x^8-9x^7+20x^6-33x^5+46x^4-66x^3+80x^2-72x+32=0$
= $2(x^8+ 16) -9(x^7+8x) +20(x^6+ 4x^2) -33(x^5+2x^3) + 46x^4=0$
or deviding by $x^4$ as x = 0 is not a solution we get
$2(x^4 + (\dfrac{2}{x})^4)-9(x^3 + (\dfrac{2}{x})^3)+20((x^2 + (\dfrac{2}{x})^2)-33(x (+\dfrac{2}{x}))+46 = 0$

now if we get $x+\dfrac{2}{x}=t$
we get
$x^2+(\dfrac{2}{x})^2=t^2-4$
$x^3+(\dfrac{2}{x})^3=t^3-6t$
$x^4+(\dfrac{2}{x})^4=t^4-8t + 8$

so given relation reduces to

$2(t^2-8t^2 +8) -9(t^3-6t) +20(t^2-4) - 33t + 46= 0$
or $2t^4-9t^3+4t^2+21t-18=0$
now we see that t = 1 and t = 3 are solutions and hence we get

$2t^4-9t^3+4t^2+21t-18=0$
= $2t^3(t-1) - 7t^2(t-1) -3t(t-1) + 18(t-1)=0$
or$(t-1)(2t^3-7t^2-3t+18) = 0$

gives a solution t = 1

or

$2t^3-7t^2- 3t + 18 = 0$ as 3 is a root we get
$2t^2(t-3) - t(t-3) - 6(t-3) = 0$
or $(t-3)(2t^2 - t^2-3) = 0$
so t = 3
or $2t^2 - t - 3 = 0 $
or $(2t-3)(t+1) = 0$
so t = 1 or 3 or - 1 or $-\dfrac{3}{2}$
now $t = x+ \dfrac{2}{x}$ and if x is positive then by AM GM inequality lowest value = $2\sqrt{2}$

or only possible value from above is
t = 3 (as t cannot be between -$2\sqrt{2}$ and $2\sqrt{2}$)
t = 3 gives x = 1 or 2 and so sum of real roots = 3

 

[anemone]

Thanks to both for participating and thanks too for posting the great solution to this challenge!