How do you solve x^4-2x^2+1=0?

Question

how would you go about solving a problem like this. I have the answer but that doesnt help me.

Answer 1

Start by replacing k = x^2

So you have k^2 - 2k + 1. This can easily be factored.
(k - 1)(k - 1) = 0

Now substitute x^2 back in.
(x^2 - 1)(x^2 - 1) = 0

You could stop here because you have (x^2 - 1) = 0
x^2 = 1
x = sqrt(1)
x = +/-1
x = -1 or x = 1

Alternatively, continue factoring:
(x - 1)(x + 1)(x - 1)(x + 1) = 0

Which will also give you the same answers:
x = 1 or x = -1.

Answer 2

Since you have only even powers of x, make the substitution y = x^2. Your equation becomes y^2-2y+1=0. This is a simple quadratic which can be factored, ((y-1)^2 solve for y, then x:

y=1, y=1 is the solution to the y equation

x^2 = y, so x=1,1,-1,-1.

Answer 3

You have 5 answers so far, and they're all correct.
But gp4rts's answer is a bit preferable to the others, because he notes that both 1 and -1 are double roots.

Because this is a quartic (4th degree) equation, it has 4 roots.
Roots can be repeated (as in this case), or they can be complex, so there may not be 4 distinct, real roots. But a quartic always has 4 roots.

Answer 4

x^4-2x^2+1=0
let x^2 be y
so equation is : y^2 - 2y +1=0
y={2 _+( squareroot of [ 4-4])}/2

y= 2/2 = 1
x^2=1
x= +_1

Answer 5

x^4-2x^2+1 = 0
(x^2 - 1)^2 = 0
x^2 - 1 = 0
x^2 = 1
x = 1 or x = -1

Th

Answer 6

let y=x^2
so you have y^2-2y+1=0
solve that using factoring and you get:
(y-1)(y-1)=0
now plug the x^2 back in for y
(x^2-1)(x^2-1)=0
factor again:
(x+1)(x-1)(x+1)(x-1)=0
answer: 1 or -1

Hope that helps!