What are the real and complex solutions of the polynomial equation?
0 = x4 + 3x2 – 4
To find the solutions of the polynomial equation 0 = x^4 + 3x^2 - 4, we can set the equation equal to zero and solve for x.
x^4 + 3x^2 - 4 = 0
Let's make a substitution to simplify the equation. We can let y = x^2.
Then, the equation becomes:
y^2 + 3y - 4 = 0
Factoring this quadratic equation, we have:
(y + 4)(y - 1) = 0
Setting each factor equal to zero, we can solve for y:
y + 4 = 0 -> y = -4
y - 1 = 0 -> y = 1
Now, we substitute back x^2 for y:
x^2 = -4
x^2 = 1
Taking the square root of both sides, we get the following solutions:
For x^2 = -4, there are no real solutions. However, there are complex solutions:
x = √(-4) = ±2i
For x^2 = 1, there are two real solutions:
x = √1 = 1
x = -√1 = -1
Therefore, the solutions to the polynomial equation 0 = x^4 + 3x^2 - 4 are:
Real solutions: x = 1, x = -1
Complex solutions: x = 2i, x = -2i