X6 plus 3x4-x2-3 equals 0

Answer 1

x6 + 3x4 - x2 - 3 = 0
(x6 + 3x4) - (x2 + 3) = 0
x4(x2 + 3) - (x2 + 3) = 0
(x2 + 3)(x4 - 1) = 0
(x2 + 3)[(x2)2 - 12] = 0
(x2 + 3)(x2 + 1)(x2 - 1) = 0
(x2 + 3)(x2 + 1)(x + 1)(x - 1) = 0
x2 + 3 = 0 or x2 + 1 = 0 or x + 1 = 0 or x - 1 = 0

x2 + 3 = 0
x2 = -3
x = ±√-3 = ±i√3 ≈ ±1.7i

x2 + 1 = 0
x2 = -1
x = ±√-1 = ±i√1 ≈ ±i

x + 1 = 0
x = -1

x - 1 = 0
x = 1

The solutions are x = ±1, ±i, ±1.7i.