Find all complex roots of $T^4-{1/2}T^2-\sqrt{15}T+{69/16}$

Question

I want to find all complex roots of $T^4-{1/2}T^2-\sqrt{15}T+{69/16}$.

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The only way I can think to do it is to find 1 complex root, $\alpha$, by inspection, so we can rearrange the polynomial to be $(T-\alpha)(T^3+pT^2+qt+r) $then show the cubic equation is irreducible by Eisenstein's criterion. But I do not think this is going to work, so any ideas would be greatly appreciated

Answer 1

It is better, I think, to try with $x^4-2x^2-8\sqrt{15}\space x+69$ where $x=2T$.

By some tedious calculation of undetermined coefficients we have

$x^4-2x^2-8\sqrt{15}\space x+69=[x^2-2\sqrt {5}\space x +(9-2\sqrt 3)]\cdot[x^2+2\sqrt {5}\space x +(9+2\sqrt 3)]$

Hence $x_{1,2}=\sqrt 5\pm\sqrt{4-2\sqrt 3}\space i$ and $x_{3,4}=-\sqrt 5\pm \sqrt{4+2\sqrt 3}\space i$

Thus $$\color{red}{\begin{cases}2T_{1,2}=\sqrt 5\pm\sqrt{4-2\sqrt 3}\space i\\ 2T_{3,4}=-\sqrt 5\pm \sqrt{4+2\sqrt 3}\space i\end{cases}}$$ The verification can be made by

$[(x-\sqrt 5)^2+(\sqrt 3-1)^2][ x+\sqrt 5)^2+(\sqrt 3+1)^2]=x^4-2x^2-8\sqrt{15}\space x+69$.