find all real and imaginary solutions -2x^4+2x^2+4=0

Answer 1

To find the real and imaginary solutions of the equation -2x^4 + 2x^2 + 4 = 0, we can use the quadratic formula by treating x^2 as a single variable.

Let's denote x^2 as y:
-2y^2 + 2y + 4 = 0

Now, we can use the quadratic formula to find the solutions for y:

y = (-b ± √(b^2 - 4ac)) / 2a

a = -2, b = 2, c = 4

y = (-2 ± √(2^2 - 4*(-2)*4)) / 2*(-2)
y = (-2 ± √(4 + 32)) / -4
y = (-2 ± √36) / -4
y = (-2 ± 6) / -4

This gives two possible solutions for y:
y1 = (6 - 2) / -4 = 4 / -4 = -1
y2 = (-6 - 2) / -4 = -8 / -4 = 2

Now, we can substitute back y = x^2 to find the values of x:

For y1 = -1:
x^2 = -1
x = ± √(-1)
x = ± i

For y2 = 2:
x^2 = 2
x = ± √2

Therefore, the solutions for the equation -2x^4 + 2x^2 + 4 = 0 are:
x = ± i, ± √2