Question

How do I find all real and complex zeros of `x^4 - 15x^2 - 75`?

Answer 1

Finding the zeros of a bi-quadratic equation is rather easy if you know how to solve the quadratics. It's all about replacing `x^2` by `y`.
Let `y = x^2`
`-> x^4−15x^2−75` `=` `y^2 - 15y - 75`

Now you can use whatever method you like to find the zeros.
`y_1 = 5/2(3+sqrt(21))` [positive]
`y_2= 5/2(3-sqrt(21))` [negative]//

For `y=x²`:
`x_(1,1) = sqrt(y_1)` and `x_(1,2) = -sqrt(y_1)`
`x_(2,1) = sqrt(y_2)` and `x_(2,2) = -sqrt(y_2)`

For `y_1` is positive and `y_2` is negative, `x_(1,1)` and `x_(1,2)` are real, but `x_(2,1)` and `x_(2,2)` are complex.

Hope it helps