find all real and complex solutions -2x^4+2x^2+4=0
To find the real and complex solutions of the equation -2x^4 + 2x^2 + 4 = 0, we can treat it as a quadratic equation in terms of x^2.
Let y = x^2, then the equation becomes -2y^2 + 2y + 4 = 0.
Now, we can solve this quadratic equation using the quadratic formula:
y = [-b ± √(b^2 - 4ac)] / 2a
where a = -2, b = 2, and c = 4.
y = [-(2) ± √((2)^2 - 4(-2)(4))] / 2(-2)
y = [-2 ± √(4 + 32)] / -4
y = [-2 ± √36] / -4
y = [-2 ± 6] / -4
Now we have two solutions for y:
1. y = (-2 + 6) / -4 = 4 / -4 = -1
2. y = (-2 - 6) / -4 = -8 / -4 = 2
Since y = x^2, we can find the values of x:
1. x^2 = -1
x = ±√(-1) = ±i (complex solution)
2. x^2 = 2
x = ±√2 (real solutions)
Therefore, the real solutions are x = √2 and x = -√2, and the complex solutions are x = i and x = -i.