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

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- July 30th 2009, 04:33 AMpacmansolve for x
Solve for x: x^4 + x^3 + x^2 + x + 1 = 0.

- July 30th 2009, 05:58 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 - July 31st 2009, 08:18 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. - July 31st 2009, 09:05 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:

01 - July 31st 2009, 09:18 PMTheAbstractionist
- July 31st 2009, 09:38 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 . . . - July 31st 2009, 10:00 PMVonNemo19
- July 31st 2009, 10:43 PMyeongil
I think there's a typo in pacman's last post. Instead of

it should have been

01 - August 10th 2009, 08:13 AMpacman
i mixed it up in a bowl of soup,

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