solve x^4-2x^2+1=0?

Question

math 112

Answer 1

Make the substitution y = x^2, reducing this to y^2 - 2y + 1 = 0. Does that look a lot easier? :)

Answer 2

First: factor the expression - multiply the 1st and 3rd coefficient to get 1. Find two numbers that give you 1 when multiplied and -2 (2nd-middle coefficient) when added/subtracted. The numbers are -1, -1

Sec: rewrite the expression with the new middle numbers... when you have 4 terms - group "like" terms...

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

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

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

(x^2 - 1)^2 = 0

Third: eliminate the exponent >find the square root of both sides...

V`(x^2 - 1)^2 = V`0
x^2 - 1 = 0

*Add 1 to both sides...

x^2 - 1 + 1 = 0 + 1
x^2 = 1

*Square both sides...

V`x^2 = +/- V`1
x = -1, 1

Answer 3

solve x^4-2x^2+1=0?
Put x^2=y, then the above Eq. becomes:
y^2 -2y+1=0 or (y-1)^2=0 which leads to y=1
or x^2=1 or
x= +1 or -1Answer.

Answer 4

x^4 - 2x^2 + 1 = 0
Factorizing
(x^2)^2 - 2. x^2 . 1 + 1^2 = 0
(x^2 - 1)^2 = 0
[(x + 1)(x - 1)]^2 = 0
Solve for x
x = -1, -1, 1, 1

Answer 5

t = x^2 , t>=0
t^2 - 2t + 1=0
<=> t = 1
<=> x^2 = 1
<=> x = +-1

Answer 6

Here is my complete solution
(x^2-1)(x^2+1) = 0
(x-1)(x+1)(x^2+1) = 0
x = 1 or x = 1

Answer 7

x^4-2x²+1=0 ==> x^4 = y²

y² - 2y + 1 = 0
d = -2² - 4.1.1
d = 4 - 4
d = 0

y = (2 +/- 0) : 2
y' = y" = 2:2 = 1
y"' = y"" = \/1 = 1
Answer:
x^4 = 1
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Answer 8

unsolvable because 2+1 not = 0
if it could be 0 then anything to the 0 power is 1 and x to the 1st power is x

Answer 9

x={1,-1}

Answer 10

the farthest i got was:
(x*x*x*x)-(2x*2x)=1