Question

How do you find the zeros of `y=9x^4+5x^2-4`?

Answer 1

`+-i and +-2/3`.

Explanation:

Solving this quadratic in `x^2`,

`x^2=-1 and 4/9`. So,

`x = +-i and +-2/3`.

See graphical depiction for real zeros, as x-intercepts `+-2/3`.

graph{9x^4+5x^2-4 [-.8, .8, -1.25, 1.25]}