This is a question from an undergraduate course on Galois theory:
Find all complex numbers which are roots of $P(T)=T^4+2T^2-\sqrt{6}T+\frac{3}{4}$
Can we use Galois theory to solves this?
Or do we just do soem long calculations to get quadratic polynomials $A(T)$ and $B(T)$ such that $P(T)=A(T)B(T)$ and then use quadratic formula?
$P(T)=T^4+2T^2-\sqrt{6}T+\frac{3}{4}=(T^2+aT+b)(T^2+cT+d)=T^4+(c+a)T^3+(d+ac+b)T^2+(ad+bc)T+bd$
$\implies $
$c+a=0 \implies a=-c$
$ d+ac+b=2 \implies d-c^2+b=2 \implies c=\sqrt{d+b-2}$
$ad+bc=-\sqrt{6} \implies -d\sqrt{d+b-2}+b\sqrt{d+b-2}=-\sqrt{6}$
$bd=\frac{3}{4} \implies b=\frac{3}{4d}$
This leads to: $-d\sqrt{d+\frac{3}{4d}-2}+\frac{3}{4d}\sqrt{d+\frac{3}{4d}-2}=-\sqrt{6}=(\frac{3}{4d}-d)\sqrt{d+\frac{3}{4d}-2}$
Squaring both sides gives: $(\frac{3}{4d}-d)^2({d+\frac{3}{4d}-2})=6$
Then we have a cubic in $d$ which I can not solve
So this would us a value of $d$ and , in turn, a value of $b$, $c$ and $a$, then we could use the quadratic formula to find complex roots.
This is very lengthy (not ideal under exam timed conditions)
Is there a more efficient way to solve this problem?