Square sum of two complex roots of $x^4-2x^3-3x-2=0$ When $\alpha$ and $\beta$ are two complex roots of
$$x^4-2x^3-3x-2=0$$
What is the value of $\alpha^2+\beta^2$?
(a) $-1\quad$
(b) $-2\quad$
(c) $-3\quad$
(d) $-4\quad$
(e) $-5\quad$

This looked like an easy question but I couldn't solve it. I tried to split the quartic polynomial into two quadratic polynomials but that was not successful either. It's surprising that the answer can be an integer. There must be some clever way to get the answer.

This is not a trivial one,
Let the roots be a,b,c,d; where c and d are real, and a,b are complex conjugates.
let $a=p+iq$,
$b=p-iq$.
We are required to find $$2(p^2-q^2)$$
Notice that $$a+b+c+d=2$$,
$$\implies c+d=2-2p$$
and,
$$abcd=-2$$,
$$\implies c.d=\frac{-2}{p^2+q^2}$$
and $$ab+bc+cd+ac+ad+bd=0$$
If I put the things that I previously derived, I get
$$p^2+q^2+4p(1-p)=\frac{2}{p^2+q^2}$$
By a similar way,
$$abc+abd+bcd+acd=3$$
If I again put the things in terms of p and q, I get:
$$2(1-p).(p^2+q^2)-\frac{4p}{p^2+q^2}=3$$
Now for these equations, Wolfram generates horrible expressions, but gives only 2 real values of p and q, as p and q are real,
$$\implies p=-0.01754 ,q=+_-1.20418$$
giving $$2(p^2-q^2)=-3$$ (approx)
