Urgent: 4 degree polynomial with irrational roots

**Billy2007** · December 10th 2006, 01:45 PM

Hi

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

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

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

where $a_0 = 1$

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

Then by fundamental theorem of algebra I see that

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

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

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

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

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

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

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

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

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

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

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

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

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

This expression makes the equation

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

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

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

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

**topsquark** · December 11th 2006, 04:33 AM

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

**CaptainBlack** · December 11th 2006, 05:47 AM

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

**ThePerfectHacker** · December 11th 2006, 07:34 AM

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!!!

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