# Thread: Roots of nth order polynomials

1. ## Roots of nth order polynomials

I'm having a heck of a time with a few sections on my differential equations homework. Solving nth order polynomials is only part of each problem, and I'm having enough trouble with that. There seems to be no method for solving these other than guess and check, and when most of them end up having complex roots it feels like it's impossible.

For example:

$r^4 + 2r^2 +1 = 0$

How are you supposed to solve for the roots of this? As far as I can tell they're complex, and I don't think I can guess and check =\ Thanks for any help guys

2. $r^4 + 2r^2 + 1 = \left( {r^2 + 1} \right)^2 = 0\quad \Rightarrow \quad r = i\,\& \, - i$

3. Just factoring we get:

$(r^{2}+1)^{2}$

The roots are clearly i and -i

4. Originally Posted by griffsterb
I'm having a heck of a time with a few sections on my differential equations homework. Solving nth order polynomials is only part of each problem, and I'm having enough trouble with that. There seems to be no method for solving these other than guess and check, and when most of them end up having complex roots it feels like it's impossible.

For example:

$r^4 + 2r^2 +1 = 0$

How are you supposed to solve for the roots of this? As far as I can tell they're complex, and I don't think I can guess and check =\ Thanks for any help guys
Alternatively

Let $r^2=\psi$

So we have

$\psi^2+2\psi+1=0$

$\psi=\frac{-2\pm\sqrt{4-4(1)(1)}}{2}\Rightarrow{\psi=-1}$
$\psi=-1\Rightarrow{r^2=-1}\Rightarrow{r=\pm{i}}$