Finding $x_1x_2+x_1x_3+x_2x_4+x_3x_4$ without explicitly finding the roots of $x^4-2x^3-3x^2+4x-1=0$ 
The equation $x^4-2x^3-3x^2+4x-1=0$ has $4$ distinct real roots $x_1,x_2,x_3,x_4$ such that $x_1\lt x_2\lt x_3\lt x_4$ and product of $2$ roots is unity, then find the value of $x_1x_2+x_1x_3+x_2x_4+x_3x_4$

This question has an answer on this link but I am trying to solve it without explicitly finding the roots because the question tells us that the product of $2$ roots is unity. I want to use it.
My Approach:
Using Descartes rule, I can see that there is one negative root and three positive roots.
Also, at $x=0, 1, -1$, the value of the polynomial is negative.
Thus, $x_1\lt-1, x_4\gt1$ and $x_2,x_3$ lies between $0$ to $1$.
Thus, I am concluding that $x_2x_4=1$ and $x_1x_3=-1$ (because product of roots is $-1$)
How to conclusively reject the case $x_3x_4=1$?
For $\alpha\gt1, \beta\gt1$, $x_1=-\beta, x_2=\frac1\alpha, x_3=\frac1\beta, x_4=\alpha$
Sum of roots$=-\beta+\frac1\alpha+\frac1\beta+\alpha=2\implies\frac1\beta-\beta=2-(\alpha+\frac1\alpha)$
Sum of product of roots taken $3$ at a time$=-\frac1\alpha+\frac1\beta-\alpha-\beta=-4\implies\frac1\beta-\beta=-4+(\alpha+\frac1\alpha)$
Therefore, $\alpha+\frac1\alpha=3, \frac1\beta-\beta=-1$
Multiplying these two, $\frac\alpha\beta-\alpha\beta+\frac1{\alpha\beta}-\frac\beta\alpha=-3$
The question asks us to find $\frac\alpha\beta-\frac\beta\alpha$, that means $-3+\alpha\beta-\frac1{\alpha\beta}$
Can we conclude this approach?

I offer to substitute $-3x^2$ as $\left(-4x^2+x^2\right)$, therefore, we will have
$$
x^4-2x^3-3x^2+4x-1=0,
\\
\left(x^4-2x^3+x^2\right)-\left(4x^2-4x+1\right)=0,
\\
x^2\left(x^2-2x+1\right)-\left(2x-1\right)^{2}=0,
\\
x^{2}\left(x-1\right)^{2}-\left(2x-1\right)^{2}=0,
\\
\left(x\left(x-1\right)\right)^{2}-\left(2x-1\right)^{2}=0,
$$
$$
\left(x\left(x-1\right)\right)^{2}-\left(2x-1\right)^{2}=0,\tag{1}
$$
Relatively $(1)$ we must use the very well-known formula
$$
\left(a^2-b^2\right)=\left(a-b\right)\left(a+b\right),\tag{2}
$$
We will receive the next
$$
\begin{cases}
x^2-3x+1=0,\\
x^2+x-1=0.\tag{3}
\end{cases}
$$
Solving $(3)$ you will obtain those $4$ roots and find $x_1x_2+x_1x_3+x_2x_4+x_3x_4$.
