Finding $x_1x_2+x_1x_3+x_2x_4+x_3x_4$ without explicitly finding the roots of $x^4-2x^3-3x^2+4x-1=0$

Question

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?

Answer 1

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 $x_1x_2+x_1x_3+x_2x_4+x_3x_4$.

Answer 2

Consider $g(x) = x^4 f(1/x) = -x^{4}+4x^{3}-3x^{2}-2x+1$. Supposing if $x_3 = 1/x_4$ and $x_4 = 1/x_3$, this would mean that $x_3, x_4$ are both roots of this new equation.

Let us verify that there are only two roots of $f(x)$ and $g(x)$ in common. $f(x) + g(x)$ has no $x^4$ term and no constant term, so it is a cubic polynomial with one root being $x = 0$. This proves our claim.

However, since we already know that $x_2 x_4 = 1$, then $x_2, x_4$ are the roots of $g(x)$. Since $g(x)$ only has these two roots, both $x_3$ and $x_4$ cannot be roots of $g(x)$. Hence $x_3 x_4 \ne 1$.