Question

2x^4+19x^2+9=0

I have x=3i and x=-3i so far, but I can't find the other solutions.

Answer 1

You can factorize this:


`2x^4+19x^2+9=(2x^2+1)(x^2+9)=2\left(x^2+\frac12\right)(x^2+9)=0`

Factorizing further (over the complex numbers) yields


`2\left(x^2+\frac12\right)(x^2+9)=2\left(x+\frac i{\sqrt2}\right)\left(x-\frac i{\sqrt2}\right)(x+3i)(x-3i)=0`

So in addition to
`x=\pm3i`, the other two solutions are
`x=\pm\frac i{\sqrt2}`.