MHB Can you solve this system of equations with four variables?

[anemone]

Find all real solutions to the following system of equations:

$a+b+c+d=5$

$ab+bc+cd+da=4$

$abc+bcd+cda+dab=3$

$abcd=-1$

 

[anemone]

Solution that I saw somewhere online:

Spoiler

We're given

$a+b+c+d=5$

$ab+bc+cd+da=4$

$abc+bcd+cda+dab=3$

$abcd=-1$

Let $X=a+c$ and $Y=b+d$. Then the system of equations is equivalent to

$x+Y=5$

$XY=4$

$Xbd+Yac=3$

$(Xbd)(Yac)=-4$

The first two of these equations imply $(X,\,Y)=(1,\,4)$ and the last two give $(Xbd,\,Yac)=(4,\,-1)$.

And this yields:

[TABLE="class: grid, width: 500"]
[TR]
[TD]$X$[/TD]
[TD]$Y$[/TD]
[TD]$Xbd$[/TD]
[TD]$Yac$[/TD]
[TD]$(a,\,c)$[/TD]
[TD]$(b,\,d)$[/TD]
[/TR]
[TR]
[TD]1[/TD]
[TD]4[/TD]
[TD]4[/TD]
[TD]-1[/TD]
[TD]$\dfrac{1\pm 2}{2}$[/TD]
[TD]2[/TD]
[/TR]
[TR]
[TD]1[/TD]
[TD]4[/TD]
[TD]-1[/TD]
[TD]4[/TD]
[TD]-[/TD]
[TD]-[/TD]
[/TR]
[TR]
[TD]4[/TD]
[TD]1[/TD]
[TD]4[/TD]
[TD]-1[/TD]
[TD]-[/TD]
[TD]-[/TD]
[/TR]
[TR]
[TD]4[/TD]
[TD]1[/TD]
[TD]-1[/TD]
[TD]4[/TD]
[TD]2[/TD]
[TD]$\dfrac{1\pm 2}{2}$[/TD]
[/TR]
[/TABLE]