Verified Answer :

LammettHash

You can factorize this:

\(2x^4+19x^2+9=(2x^2+1)(x^2+9)=2\left(x^2+\dfrac12\right)(x^2+9)=0\)

Factorizing further (over the complex numbers) yields

\(2\left(x^2+\dfrac12\right)(x^2+9)=2\left(x+\dfrac i{\sqrt2}\right)\left(x-\dfrac i{\sqrt2}\right)(x+3i)(x-3i)=0\)

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

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