I have tried everything to get the 4 anwsers to this problem hope someone can help me.(2x^4)-(3x^2)+1=0?

Answer 1

This should be recognized a s a quadratic eq. in x squared. Several ways to do it. The classic way is to let u = x^2 and then the original becomes 2u^2-3u+1=0. Factor and set both factors = 0. Then replace back x^2 for u and factor yet again for 4 eq's.

Answer 2

Let z = x^2 then
2z^2 - 3z + 1 = 0
(2z-1)(z-1) = 0
2z = 1 or z = 1/2 = x^2
z = 1 or x^2 = 1
So the four answers are
x^2 = 1/2, x = +-sqr[2]/2
x^2 = 1, x = +- 1

Answer 3

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

so 2x^2 = 1;x^2 =1.
x = 1 or x = -1
x = 1/2^1/2 or x = -(1/2^/1/2)( sorry I cannot type square root)

Answer 4

Let z = x^2
Then 2z^2 -3z +1 = 0
(2z-1)(z-1) = 0
z = 1/2 and z= 1
So x^2 = 1/2 --> x = +/- sqrt(1/2)= +/- .5sqrt(2), and
x^2 = 1 --> x = +/- 1
So there's your 4 solutions

Answer 5

2x^4 - 3x^2 = -1
x^2(2x^2 - 3) = -1

-1, 1 and 1/2*sqrt(2), -1/2*sqrt(2)

Answer 6

2x^4-3x^2+1=0
put x^2=y;
2y^2-3y+1=0
2y^2-2y-y+1=0
2y(y-1)-1(y-1)=0
(y-1)(2y-1)=0
y=1,1/2.
x^2=1,1/2
x=+/-1, +/-1/sqrt(2). ANS.