What are the real and complex solutions of the polynomial equation?
0 = x4 + 3x2 – 4
To find the real and complex solutions of the given polynomial equation, we can rearrange it to the form:
x^4 + 3x^2 - 4 = 0
Let's substitute x^2 = y:
y^2 + 3y - 4 = 0
Factoring the quadratic equation:
(y + 4)(y - 1) = 0
This gives us two possible values for y:
y + 4 = 0 or y - 1 = 0
y = -4 or y = 1
Substituting back y = x^2:
Case 1: x^2 = -4
Taking the square root of both sides, considering both positive and negative roots:
x = ±√(-4)
x = ±2i
Case 2: x^2 = 1
x = ±√(1)
x = ±1
Therefore, the real solutions are x = -1 and x = 1, and the complex solutions are x = -2i and x = 2i.