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!

 

[CellCoree]

I would first like to clarify that the given equation is a polynomial with degree 8, and therefore may have up to 8 real roots. I will also assume that the coefficients are real numbers.

To find the sum of the real roots, we can use the fundamental theorem of algebra, which states that the number of complex roots of a polynomial is equal to its degree. Since the degree of the given polynomial is 8, we can conclude that it has 8 complex roots.

However, we are only interested in the sum of the real roots. In order to find this, we can use the fact that complex roots occur in conjugate pairs. This means that if a+bi is a complex root, then so is a-bi. Therefore, the sum of these two complex roots is 2a, which is a real number.

Applying this concept to all 8 complex roots, we can pair them up and find the sum of each pair. This will result in 4 real numbers, which we can then add together to find the total sum of the real roots.

In summary, the sum of the real roots for the given polynomial is equal to the sum of 4 real numbers, which can be found by pairing up the complex roots and adding them together.