Question

my head hurts, I could really use some help here! use the given root to help find the remaining roots of the equation. x^4-2x^2+1=0; 1 (double root)

Answer 1

(x^2-1)^2= 0
x^2 = 1
x = 1, -1

Answer 2

I think I was making it way too hard.

Answer 3

did u get it clearly?

Answer 4

Its a perfect square. If you let y=x^2
then you have y^2-2y+1=0
(y-1)^2=0
but y=x^2
(x^2-1)^2=0

Answer 5

the example I had broke it down into three groups... like (x-2)(x+2)(x-3) and I was trying to make this one work like that... was I going about it wrong? the above listed were just random example figures, not actual answers

Answer 6

Well, from (x^2-1)(x^2-1) you get (x+1)^2(x-1)^2

Answer 7

Let p(x) = x^4 - 2x^2 + 1 = 0
if 1 is a root of p(x), the (x-1) is a factor of p(x) ------Factor Theorem
but 1 is double root, this means two of the factors are (x-1) and (x-1)
taking their product and dividing p(x) by (x^2-2x +1), we get the quotient
as (x^2 + 2x +1) i.e. (x + 1)^2
so the other two factors are (x+1)(x+1)

so all four factors are 1,1, -1 and -1

Answer 8

thank you ssnapier

Answer 9

* sorry
all four roots are 1, 1, -1, -1

Answer 10

ok, Harkirat... you are all over it, thanks!

Answer 11

you are also welcome get free help using
: sudhanshu.cochin@gmail.com

Answer 12

u r welcome.... :)