x^4 - 5x^2 + 6 = 0 .......... what are the 4 solutions?

Answer 1

This equation is quadratic in form, that is, it can be transformed into a quadratic equation by a change in variables, and solved by the usual methods of solving quadratics.
replace y = x^2, so that x = +/- sqrt y.
Then solve
y^2 - 5y + 6 = 0..
factor and solve by the zero factor theorem
(y-3)(y-2) = 0
y = 3 or y = 2
x = +/- sqrt 3 or +/- sqrt 2
For more info see..
http://www.wtamu.edu/academic/anns/mps/math/mathlab/col_algebra/col_alg_tut20_quadform.htm
http://www.purplemath.com/modules/solvquad.htm

Answer 2

x^4 - 5x^2 + 6 = 0
(X^2 - 2)(X^2 - 3) = 0
(X^2 - 2) = 0 (X^2 - 3) = 0
X^2 = 2 X^2 = 3
X =+- SQRT(2) X = +-SQRT(3)

Answer 3

(x^2-2)(x^2-3)=0
x^2-2=0
x^2-3=0
x=±√2 and ±√3

Answer 4

(x^2-2)(x^2-3)

(x+1)(x-2)(x+1)(x-3)

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