The roots of x^8 - 5x^6 + 7x^4 - 5x^2 +6=0

[mathnovice]

[Thread moved to homework forum by a Mentor]
the exercise was to find the roots of x^8 - 5x^6 + 7x^4 - 5x^2 +6=0

I substituded x^2 with y
: y^4 - 5y^3 + 7y^2 - 5y +6=0

I factored this by doing the rational roots test and trying those possible roots with the method of horner

and got (y-2) (y-3)(y^2+1)=0

with y=x^2 ---> (x^2-2) (x^2-3)(x^4+1)=0

i found the roots ±√2 , ±√3. which is correct and the answer on the back says these are the 2 anwsers. but (x^4+1) remains --> what are the 2 complex roots? how do I change 4^√-1 in a notation with i?

 

[HallsofIvy]

$x^4+ 1= 0$ is, of course, the same as $x^4= -1$ so the solutions are the four fourth roots of -1. You can find them using DeMoivres theorem: The nth roots of $r(cos(\theta)+ i sin(\theta)$ are $r^{1/n}(cos(\frac{1}{n}(\theta+ 2k\pi)+ isin(\theta+ 2k\pi))$ where k runs from 0 to n- 1.

Here, r= 1 and $\theta= \pi$.

 

[BvU]

Ah, and welcome to PF, dear mathnovice ! Good attitude to ask a little more !It's a fourth order, so you in fact expect four solutions !

Given that $i^2 = -1$ you are left with solving $x^2 = i$ or $x^2 = -i$

Now it becomes a little weirder: one solution for the first one is $x = {1\over \sqrt 2}(1+i)$ !

Check by writing out: $x^2 = {1\over 2}(1+i)(1+i) = {1\over 2}(1^2+2i+i^2) = {1\over 2}(1+2i-1) = i$

I could go on, but maybe you would like to find the other three solutions by yourself ?

Then draw the solutions in a Cartesian coordinate system where instead of x and y you have the real part of x (so ${1\over \sqrt 2}$ in my example ) horizontally and the imaginary part of x vertically. Welcome to the world of imaginary numbers !

 

[Ray Vickson]

[Thread moved to homework forum by a Mentor]
the exercise was to find the roots of x^8 - 5x^6 + 7x^4 - 5x^2 +6=0

I substituded x^2 with y
: y^4 - 5y^3 + 7y^2 - 5y +6=0

I factored this by doing the rational roots test and trying those possible roots with the method of horner

and got (y-2) (y-3)(y^2+1)=0

with y=x^2 ---> (x^2-2) (x^2-3)(x^4+1)=0

i found the roots ±√2 , ±√3. which is correct and the answer on the back says these are the 2 anwsers. but (x^4+1) remains --> what are the 2 complex roots? how do I change 4^√-1 in a notation with i?

To amplify a bit on what others have told you: we can immediately find two 4th roots of $-1$ by noting that
$$-1 = e^{\pm i \pi} \Longrightarrow (-1)^{1/4} = e^{\pm i \pi/4} = \frac{1 \pm i}{\sqrt{2} }$$
However, if $r$ is a fourth root of $-1$ then so is $r^3$; do you see why? This gives us two other fourth roots
$$e^{\pm 3 i \pi/4} = \frac{? \pm i \,?}{?}$$
You can fill in the remaining details.

 

[BvU]

how do I change 4^√-1 in a notation with i?

With this wording, I expect the poster to be helped with an explanation at an introductory level. Dear novice, did the replies make you any wiser ?

 

[mathnovice]

Everybody thanks for the replies. I am now improving my knowledge about complex numbers first and will then tackle this problem. But it did motivate me to discover the whole subjet of complex numbers, as I just find out complex numbers are also used in engineering applications and that's my goal for improving my mathknowledge ( I'm going to study engineering at university next year )