anyone can tell me how to get the roots of this equation ill give medal...
x^6+9x^4+24x^2+16=0

Question

anyone can tell me how to get the roots of this equation ill give medal...
x^6+9x^4+24x^2+16=0

sooobored · Answer

have you tried long dividing or synthetic division factors of 16?

sooobored · Answer

oh, fyi, make the substitution
x^2 =u

it will make the wholeee problem much easier

caerus · Answer

i tried synthetic, and stocked up

sooobored · Answer

stocked up?

caerus · Answer

sooobored · Answer

ah ha, cause the answers are all imaginary

caerus · Answer

that site can do imaginary too

sooobored · Answer

can it? cause matlab gave me actual imaginary solutions

caerus · Answer

can you give me the site?

sooobored · Answer

ok, make the u substituion $$u^3 +9x^2+24u+16$$

caerus · Answer

cuz this roots, i will work for linear equations w/ constant coefficient

sooobored · Answer

matlab is a program, a higher tech calculator that engineering students generally use

sooobored · Answer

ok, note how all positives, so it must have the form (x+n)

factors of 16- 1,2,4,8,16

try synthetic division using -1

caerus · Answer

by that u substitution i get u=-1,-4,-4

caerus · Answer

what can i do next?

caerus · Answer

ill substitute the value of u in u=D^2 ?

sooobored · Answer

no, substitute back in $x^2$

sooobored · Answer

when factored it should look like 
$ (x^2 +4)(x^2 +4)(x^2 +1)$

sooobored · Answer

now,
remember how 
$ x^2-n^2=(x-n)(x+n)$

sooobored · Answer

well, $ i *i =-1$

caerus · Answer

nice, thank i got the idea of u substitution, thanks for your help

sooobored · Answer

hence, 
$x^2 +n^2= (x-in)(x+in)$

sooobored · Answer

yup, no problem