What are the real and complex solutions of the polynomial equation x^4 - 13x^2 = -36?
Solve for x over the real numbers:
x^4-13 x^2 = -36
Add 36 to both sides:
x^4-13 x^2+36 = 0
Substitute y = x^2:
y^2-13 y+36 = 0
The left hand side factors into a product with two terms:
(y-9) (y-4) = 0
Split into two equations:
y-9 = 0 or y-4 = 0
Add 9 to both sides:
y = 9 or y-4 = 0
Substitute back for y = x^2:
x^2 = 9 or y-4 = 0
Take the square root of both sides:
x = 3 or x = -3 or y-4 = 0
Add 4 to both sides:
x = 3 or x = -3 or y = 4
Substitute back for y = x^2:
x = 3 or x = -3 or x^2 = 4
Take the square root of both sides:
Answer: | x = 3 or x = -3 or x = 2 or x = -2
