What are the real and complex solutions of the polynomial equation?
x^4 -29x^2=-100
To find the solutions of the polynomial equation x^4 - 29x^2 = -100, we can rearrange the equation to bring everything to one side and then factorize it.
x^4 - 29x^2 + 100 = 0
Now, let's introduce a substitution to simplify the equation.
Let y = x^2.
Plugging this into the equation, we get:
y^2 - 29y + 100 = 0
Now, we can factorize this quadratic equation:
(y - 4)(y - 25) = 0
Setting each factor to zero, we have:
y - 4 = 0 or y - 25 = 0
Solving for y, we find:
y = 4 or y = 25
Now, we substitute back y = x^2
For y = 4:
x^2 = 4
Taking the square root on both sides:
x = ±2
For y = 25:
x^2 = 25
Taking the square root on both sides:
x = ±5
Thus, the real solutions to the equation x^4 - 29x^2 = -100 are x = -5, -2, 2, and 5.