# solve for x

• Jul 30th 2009, 03:33 AM
pacman
solve for x
Solve for x: x^4 + x^3 + x^2 + x + 1 = 0.
• Jul 30th 2009, 04:58 AM
songoku
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 PM
pacman
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 PM
yeongil
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:
$x = \cos \frac{2\pi}{5} + i\sin \frac{2\pi}{5}$
$x = \cos \frac{8\pi}{5} + i\sin \frac{8\pi}{5}$
$x = \cos \frac{4\pi}{5} + i\sin \frac{4\pi}{5}$
$x = \cos \frac{6\pi}{5} + i\sin \frac{6\pi}{5}$

01
• Jul 31st 2009, 08:18 PM
TheAbstractionist
Quote:

Originally Posted by pacman
Solve for x: x^4 + x^3 + x^2 + x + 1 = 0.

Observe that $(x-1)(x^4+x^3+x^2+x+1)=x^5-1.$ Hence the solution you want is the set of the four complex 5th roots of unity. No need to plot graph or use computer.
• Jul 31st 2009, 08:38 PM
pacman
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 PM
VonNemo19
Quote:

Originally Posted by pacman
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 . . .

$x+\frac{1}{x}=\frac{-1\pm\sqrt{-3}}{2}$

Then

$x=\frac{-\left(\frac{-1\pm\sqrt{-3}}{2}\right)\pm\sqrt{\left(\frac{-1\pm\sqrt{-3}}{2}\right)^2-4}}{2}$

Is this what you are saying?
• Jul 31st 2009, 09:43 PM
yeongil
I think there's a typo in pacman's last post. Instead of
$\left(x + \frac{1}{x}\right)^2 + \left(x + \frac{1}{x}\right) + 1 = 0$
it should have been
$\left(x + \frac{1}{x}\right)^2 + \left(x + \frac{1}{x}\right) {\color{red}-}\; 1 = 0$

01
• Aug 10th 2009, 07:13 AM
pacman
i mixed it up in a bowl of soup, yeongil is correct with his observation . . . .