Solving an equation in 4th degree

[zorro]

Homework Statement

How to solve x⁴ + x³ + x² + x + 1= 0 ?

 

[The legend]

Any 4th degree eqn can be solved by Ferrari's method
http://www.proofwiki.org/wiki/Ferrari's_Method

by the way ... how did you attempt it?
Did you try the "substitute some random value" method?

 

[zorro]

Yes, I tried it but it did not work. Ferrari's method looks good.

 

[The legend]

it IS good! (sure shot solution guaranteed)

 

[HallsofIvy]

$x^4+ x^3+ x^2+ x+ 1$ is a "cyclotomic" polynomial whose zeros are equally spaced around the unit circle in the complex plane.

One way to solve this is to convert it to a 5th degree equation (for which there is no general formula!). Multiplying $x^4+ x^3+ x^2+ x+ 1= 0$ by x- 1 gives $(x- 1)(x^4+ x^3+ x^2+ x+ 1)= (x^5+ x^4+ x^3+ x)$$- (x^4+ x^3+ x+ 1)= x^5- 1= 0$.

Now, the roots of $x^5- 1= 0$ are the complex numbers $e^{2\pi ki/5}$ where k goes from 0 to 4. Obviously, k= 0 gives $x= 1$ which statisfies x- 1= 0 but not $x^4+ x^3+ x^2+ 1= 0$ so the roots of $x^4+ x^3+ x^2+ x+ 1$ are
$e^{2\pi ki/5}= cos(\frac{2\pi k}{5})+ i sin(\frac{2\pi k}{5})$
for k= 1, 2, 3, and 4.

 

[awkward]

I like HallsOfIvy's solution, but just for variety here is another approach.

Divide by $$x^2$$, yielding

$$x^2 + x + 1 + x^{-1} + x^{-2} = 0$$.

Now let $$y = x + x^{-1}$$.

With this substitution, the equation becomes

$$y^2 + y - 1 = 0$$.

Solve for y (two roots), then solve for x (four roots).

 

[The legend]

nice one awkward...HallsofIvy has an nice solution too(one i never thought of)...

 

[zorro]

Divide by LaTeX Code: x^2 , yielding

LaTeX Code: x^2 + x + 1 + x^{-1} + x^{-2} = 0 .

Now let LaTeX Code: y = x + x^{-1} .

With this substitution, the equation becomes

LaTeX Code: y^2 + y - 1 = 0 .

Solve for y (two roots), then solve for x (four roots).

That is a very ingenious solution. Thanks.