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

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Answer 1 · 2017-01-16T12:20:58

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

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]}

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