Can anyone solve x^8+x^5+1=0?

Question

Sorryyyy it's not +1 it's -1 at the end. It should have real roots now...

Answer 1

♠ LOL! You little crow! Look how much fuss you’ve done to other guys!
♠ Ok; y(x)=x^8+x^5-1; first let’s sniff it around;
y’= 8x^7 +5x^4 =0, hence x^4*(8x^3+5)=0,
hence x1=0, x2= -(5/8)^(1/3) could be suspected as local extremums;
y’’= 56x^6 +20x^3, hence y’’(x1) =0 means no extremum,
and y’’(x2) =-(5/8)*(56*(-5/8) +20) > 0 means local min;
♦ and y(x2) < 0 means there are 2 intercepts!
Now notorious Newton says x[j+1] = x[j] –y(x[j])/y’(x[j]);
1) Start with x[1] =x2-1.2 ≈ -2, then
x[2] = -2 –(2^8 -2^5 –1)/(-8*2^7 +5*2^4 ) =-1.764;
x[3] = -1.563; x[4]=-1.397; ;;;;
x[9]= -1.1461390; x[10]= -1.1461390;
2) Start with x[1] =x2+3 ≈ 2, then
x[2] = 2 –(2^8 +2^5 –1)/(8*2^7 +5*2^4 ) =1.740;
x[3] = 1.511; x[4]=1.311; ;;;;
x[10]= 0.8970237; x[11]= 0.8970237;

Answer 2

That equation has no real solutions. Check the graph of x^8 + x^5+1 here:
http://math.colgate.edu/~kellen/interspace/nosolns.gif

It also cannot be solved in any nice way because its roots cannot be written in a closed elementary form. What does that mean?

For liner and quadratic equations, you know how to solve for when f(x) = 0. Lines are simple, and there's a formula for quadratic ones. (The quadratic formula.) There is a formula for 3rd degree equations, and even 4th, but after that, there's no equation, and for most polynomials of degree 5 or higher, there's no way to write the solutions down. At all. No combination of +s, -s, fractions, or radicals will give you an expression for the roots. You have to approximate. See:
http://en.wikipedia.org/wiki/Galois_theory#Solvable_groups_and_solution_by_radicals
http://en.wikipedia.org/wiki/Abel-Ruffini

We do know, however, that there are 8 complex roots that we can approximate. They are as follows:

-1.00377 - 0.270225 i

-1.00377 + 0.270225 i

-0.314537 - 0.844821 i

-0.314537 + 0.844821 i

0.490558 - 1.02215 i

0.490558 + 1.02215 i

0.827745 - 0.448046 i

0.827745 + 0.448046 i

Answer 3

Take y = x^8+x^5+1
y´= 8x^7+5x^4
y´=0 x^4(8x^3+5)=0 x=0 and x= -0.8550 a relative minimum
f(-0.8550)=0.2885-0.4569+1>0.
As lim y for x==>+-infinity is +infinity the equation has NO real roots

Answer 4

Sorry...if you had something like x^10 + x^5 + 1 = 0, you could substitute y=x^5 and solve for a quadratic in y, then work backwards to get x.

...but you need someone smarter than me here.

Answer 5

This problem is so easy, it practically solves itself. ;-)


{{x -> -1.00377 - 0.270225 i},
{x -> -1.00377 + 0.270225 i},
{x -> -0.314537 - 0.844821 i},
{x -> -0.314537 + 0.844821 i},
{x -> 0.490558- 1.02215 i},
{x -> 0.490558+ 1.02215 i},
{x -> 0.827745- 0.448046 i},
{x -> 0.827745+ 0.448046 i}}