# Urgent: 4 degree polynomial with irrational roots

• Dec 10th 2006, 01:45 PM
Billy2007
Urgent: 4 degree polynomial with irrational roots
Hi

which specific formula/method I can use to find the complex roots of a polynomial of degree 4 on form:

$\displaystyle p(x) = x^4 + a x^3 + bx^2 + ax + a_0 = 0$

which has irrational coefficients
a = $\displaystyle -1 - \sqrt(2)$ and b = $\displaystyle 2+ \sqrt(2)$ and

where $\displaystyle a_0 = 1$

If I write the equation as $\displaystyle x^4-x^3+2x^2-x+1-\sqrt2(x^3-x^2+x)=0$, then you will notice that it factorises as $\displaystyle (x^2-x+1)(x^2-\sqrt2x+1)=0$

Then by fundamental theorem of algebra I see that

$\displaystyle p(x) = (x-(\frac{1}{2}) + \frac{\sqrt(3)}{2}i)) (x-(\frac{1}{2}) - \frac{\sqrt(3)}{2}i)) \cdot ((x - (\frac{\sqrt(2)}{2} + \frac{\sqrt(2)}{2} i)) ((x - (\frac{\sqrt(2)}{2} - \frac{\sqrt(2)}{2} i)) = (x^2-x+1)(x^2-\sqrt2x+1) = x^4-x^3+2x^2-x+1-\sqrt2(x^3-x^2+x)$

where the complex roots of the original polymial p(x) is

$\displaystyle x = ((\frac{1}{2}) \pm \frac{\sqrt(3)}{2}i), (\frac{\sqrt(2)}{2} \pm \frac{\sqrt(2)}{2} i)$

Getting back to the original polynomial
(*)$\displaystyle p(x) = x^4 + ax^3 + bx^ 2 + ax + 1 = 0$

where $\displaystyle a,b \in \mathbb{C}$

I would like to prove that a complex number x makes (*) true iff

$\displaystyle s = x + x^{-1}$ is a root of the $\displaystyle Q(s) = s^2 + as + (b-2)$

I see that that $\displaystyle Q(x + x^{-1}) = \frac{p(x)}{x^2}$

So the solutions of in my equation $\displaystyle s^2+as+(b-2)=0$ are $\displaystyle s=\frac{-a\pm\sqrt{a^2-4(b-2)}}{2}$.

I then need to plug each of these numbers into the equation $\displaystyle s=x+x^{-1}$, which can be written as $\displaystyle x^2-sx+1=0$. Using the quadratic formula again, you get $\displaystyle x=\frac{s\pm\sqrt{s^2-4}}{2}$.

By the way do x^-1 then exist ?
Yes, there's no problem about that. In fact since $\displaystyle s=x+x^{-1}$, you can see that $\displaystyle x^{-1}=s-x$

I get this x value for both values of s to be:

$\displaystyle x =\frac{\sqrt{-2(a \sqrt{a^2 - 4 \cdot (b-2) - a^2 + 2(b+2)}} + \sqrt{a^2 - 4 (b-2) -a}}{4}$

This expression makes the equation

$\displaystyle p(x) = \frac{p(x)}{x^2}$ true

since $\displaystyle p(\frac{\sqrt{-2(a \sqrt{a^2 - 4 \cdot (b-2) - a^2 + 2(b+2)}} + \sqrt{a^2 - 4 (b-2) -a}}{4}) = \newline \frac{p(\frac{\sqrt{-2(a \sqrt{a^2 - 4 \cdot (b-2) - a^2 + 2(b+2)}} + \sqrt{a^2 - 4 (b-2) -a}}{4})}{(\frac{\sqrt{-2(a \sqrt{a^2 - 4 \cdot (b-2) - a^2 + 2(b+2)}} + \sqrt{a^2 - 4 (b-2) -a}}{4})^2} = 0$

This proves that there exist a number x which is both a root of the original polynomial and Q(s = x + x^{-1}), since $\displaystyle p(x) = \frac{p(x)}{x^2}$.

But one question remains on my part. Since x is supposedly a complex number. Is it a complex number in its current form? That I'm a bit unsure of. Since then it should be written in the form x = a + bi ?

Best Regards,
Billy
• Dec 11th 2006, 04:33 AM
topsquark
If you want to call x a complex number, it's a complex number. There's no need to write it in a + bi form, though you may do so if you like.

-Dan
• Dec 11th 2006, 05:47 AM
CaptainBlack
You could try running the particular form of quartic that you have through
the quartic formula and see what you get.

Background on the formula can be found here.

RonL
• Dec 11th 2006, 07:34 AM
ThePerfectHacker
I know how to solve the cubic algebraically but never learned how to solve the quartic, it is too long. I find it amazing how these Italian mathematicians were so good at manipulation of radical terms and factorizations.

I read that one Italian mathematician was killed for being able to solve the quartic. Because people believed it was not withing human skills and thus he made a pact with a Devil (similar story with the violinist, Pagannini, they used the same excuse).

This is my 37:):)th Post!!!