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?

For $ \ f(x) \ = \ x^4 - 2x^3 - 3x^2 + 4x-1 \ \ , \ $ the "depressed" polynomial is $ \ f \left(x + \frac12 \right) $ $ \ = \ f(y) \ = \ y^4 - \frac92·y^2 + \frac{1}{16} \ \ , \ $ which, as an even function, has symmetrically-arranged zeroes given by $ \ y^2 \ = \ \frac14·( \ 9 \pm \sqrt{80} \ ) \ \ . \ $  We can approximate the locations as
$$   y^2 \ \ = \ \  \frac14·\left( \ 9 \ \pm \ 9·\sqrt{1 \ - \ \frac{1}{81}} \ \right) \ \ \approx \ \ \frac94·\left( \ 1 \ \pm \ \left[ \ 1 \ - \ \frac{1}{2·81} \ \right] \ \right) \ \ \approx \ \ \frac{1}{8·9} \ \ , \ \ \frac92 \ \ . $$
The four zeroes of $ \ f(x) \ $ are thus estimated by
$$ x_1 \ \approx \ \frac12  -  \frac{3}{\sqrt2} \ < \ 0 \ \ \ , \ \ \ x_2 \ \approx \ \frac12  -  \frac{1}{6\sqrt2}  \ \ \ , \ \ \ x_3 \ \approx \ \frac12  +  \frac{1}{6\sqrt2}  \ \ \ , \ \ \ x_4 \ \approx \ \frac12  +  \frac{3}{\sqrt2} \ > \ 2 \ \ . $$
What is clear from this is that $ \ x_3·x_4 \ $ cannot be equal to  $ \ 1 \ $ and that the only product of two zeroes than can be is $ \ \mathbf{x_2}·x_4 \ \ . \ $  [The exact values are in fact $ \ -\phi \ \ , \ \ 2 - \phi \ \ , \ \ \frac{1}{\phi} \ = \ \phi - 1 \ \ , \ \  \phi + 1 \ \ , \ $ but we don't need to know that in order to resolve this particular issue.]
To return to the main question, if we label the four zeroes of $ \ f \left(x + \frac12 \right) \ $  as $ \ -\gamma \  ,  \ -\delta \  ,  \ +\delta \  ,  \ +\gamma \ \ , \ $ then their product is $ \ \gamma^2·\delta^2 \ = \ \frac{1}{16} \ \Rightarrow \ \gamma·\delta \ = \ \frac14 \ \ . \ $  (The estimates above actually happen to fit this nicely.)  We may then express the zeroes of $ \ f(x) \ $ as (using your notation)
$$ x_1 \ \ = \ \ \frac12 \ - \ \gamma \ \ = \ \ -\beta \ \ \ , \ \ \ x_2 \ \ = \ \ \frac12 \ - \ \frac{1}{4 \ \gamma} \ \ = \ \ \frac{1}{\alpha}  \ \ \ , $$ $$ x_3 \ \ = \ \ \frac12 \ + \ \frac{1}{4 \ \gamma} \ \ = \ \ \frac{1}{\beta}  \ \ \ , \ \ \ x_4 \ \ = \ \ \frac12 \ + \ \gamma \ \ = \ \ \alpha \ \ . $$
We can establish from these results that
$$ \alpha·\frac{1}{\alpha} \ \ = \ \ \left( \ \gamma \ + \ \frac12 \ \right)·\left( \ \frac12 \ - \ \frac{1}{4 \ \gamma} \ \right) \ \ = \ \  \frac{4·\gamma^2 \ - \ 1}{8 \ \gamma} \ \ = \ \ 1 $$
$$ \Rightarrow \ \ 4·\gamma^2 \ - \ 8·\gamma \ - \ 1 \ \ = \ \ 0 \ \ \Rightarrow \ \ \gamma \ \ = \ \ \frac{8 \ + \ \sqrt{64 \ + \ 16}}{8} \ \ = \ \ 1 \ + \ \frac{  \sqrt5}{2} \ \ ,  $$
where we have retained only the positive  solution (we obtain the same result from $ \ \beta·\frac{1}{\beta} \ \ ) $
$$ \Rightarrow  \ \ \alpha \ \ = \ \ \frac{3 +  \sqrt5}{2} \ \ ( \ = \ x_4 \ ) \ \ \ , \ \ \ \beta \ \ = \ \ \frac{1 +  \sqrt5}{2} \ \ ( \ = \ -x_1 \ ) \ \ $$
$$ \Rightarrow  \ \ \alpha·\beta \ \ = \ \ \frac{8 \ + \  4\sqrt5}{4} \ \ = \ \ \ 2 \ + \ \sqrt5 \ \   \Rightarrow \ \ \frac{1}{\alpha·\beta} \ \ = \ \ \frac{  2 \ - \ \sqrt5  }{4 \ - \ 5} \ \ = \ \ \sqrt5 \ - \ 2  $$
$$ \Rightarrow \ \ \alpha·\beta \ - \ \frac{1}{\alpha·\beta} \ \ = \ \ 4 \ \ \Rightarrow \ \  x_1  x_2 \ + \ x_1  x_3 \ + \ x_2  x_4 \ + \ x_3  x_4 \ \ = \ \  -3 \ + \ 4 \ \ = \ \ 1 \ \ . $$
[Using the exact value of the zeroes written above in terms of the Golden Ratio $ \ \phi \ \ , \ $ we verify that
$$ (-2 \phi \ + \ \phi^2) \ + \ (-\phi^2 \ + \ \phi) \ + \ (2 \phi \ - \ \phi^2 \ + \ 2 \ - \ \phi) \ + \ (\phi^2 \ - \ 1) \ \ = \ \ 1 \ \ . \ ] $$
So while we have succeeded in finding the value of the sum of the specified pair-products of zeroes without explicitly determining all of the zeroes, it appears to be necessary to characterize those zeroes to a certain extent to evaluate the sum by your approach.
