Solve for x: x^4 + x^3 + x^2 + x + 1 = 0.

Printable View

- Jul 30th 2009, 03:33 AMpacmansolve for x
Solve for x: x^4 + x^3 + x^2 + x + 1 = 0.

- Jul 30th 2009, 04:58 AMsongoku
If you draw the graph of x^4 + x^3 + x^2 + x + 1, it won't cut the x-axis.

So, x^4 + x^3 + x^2 + x + 1 = 0 has no real roots - Jul 31st 2009, 07:18 PMpacman
Thanks

**songoku**, (Rock)(Rock)(Rock) I have an UGLY solution for this, but i am hoping THAT someone may be able to solve this elegantly. - Jul 31st 2009, 08:05 PMyeongil
If anyone's curious, the four complex roots are the following:

x ≈ 0.309 + 0.951i

x ≈ 0.309 - 0.951i

x ≈ -0.809 + 0.588i

x ≈ -0.809 - 0.588i

(using a computer, of course (Wink))

Polar form:

$\displaystyle x = \cos \frac{2\pi}{5} + i\sin \frac{2\pi}{5}$

$\displaystyle x = \cos \frac{8\pi}{5} + i\sin \frac{8\pi}{5}$

$\displaystyle x = \cos \frac{4\pi}{5} + i\sin \frac{4\pi}{5}$

$\displaystyle x = \cos \frac{6\pi}{5} + i\sin \frac{6\pi}{5}$

01 - Jul 31st 2009, 08:18 PMTheAbstractionist
- Jul 31st 2009, 08:38 PMpacman
Hint: divide BS x^2, we have

x^2 + x + 1 + 1/x + 1/x^2 = 0, rearranging it and then grouping it

(x + 1/x)^2 + (x + 1/x) + 1 = 0

It is now quadratic in x + 1/x, the result is quite ugly but it can generate a result . . . - Jul 31st 2009, 09:00 PMVonNemo19
- Jul 31st 2009, 09:43 PMyeongil
I think there's a typo in pacman's last post. Instead of

$\displaystyle \left(x + \frac{1}{x}\right)^2 + \left(x + \frac{1}{x}\right) + 1 = 0$

it should have been

$\displaystyle \left(x + \frac{1}{x}\right)^2 + \left(x + \frac{1}{x}\right) {\color{red}-}\; 1 = 0$

01 - Aug 10th 2009, 07:13 AMpacman
i mixed it up in a bowl of soup,

**yeongil**is correct with his observation . . . .