Solving the system $b+c+d=4$, $ad+bc=-8$, $a+b=5$, $cd=-8$

Question

There is a given system of equations
begin{align} ab+c+d&=phantom{-}4 \ ad+bc&=-8 \ a+b&=phantom{-}5 \ cd&=-8 end{align}
I have tried to simplify it by multiplying one equation by another or adding one to another or similar. Nothing came up.
Any hints would be great! Thank you.

Somos · Accepted Answer

From the 3rd and 4th equations we find that
$$b=5-aquad text{ and }quad d=-8/c. tag{1}$$
Substituting these in the 1st and 2nd equations they become
$$ x:=(c^2+5ac-a^2c-4c-8)/c tag{2} = 0$$ and
$$ y:=(5c^2-ac^2-8a+8c)/c = 0. tag{3}$$
The polynomial resultant of $,xcdot c,$ and $,ycdot c,$
eliminating $,c,$ is
$$ 8(a-2)(a-3)(a^2-4a-1)(a^2-6a+4). tag{4}$$
This has six roots for $,a$. For each value of $,a,$
the GCD of $,x,$ and $,y,$ uniquely determines $,c,$ and the values for $,b,$ and
$,d,$ are uniquely determined from equations in $(1)$.
Of course, there is nothing unique about picking the 3rd and 4th
equations and solving for $b$ and $d$. Also, I decided to use resultants
to solve for $a$, but there are alternative ways to do so.

g.kov · Answer

begin{align} ab+c+d&=-4 tag{1}label{1} ,\ ad+bc&=-8 tag{2}label{2},\ a+b&=-5 tag{3}label{3},\ cd&=-8 tag{4}label{4}.end{align}

Substitution of
begin{align}
a&=5-b
tag{5}label{5}
,\
d&=-frac8c
tag{6}label{6}
end{align}
into eqref{2} results in
begin{align}
b&=frac{40-8c}{c^2+8}
tag{7}label{7}
,\
a &= frac{c(5c+8)}{c^2+8}
tag{8}label{8}
.
end{align}
Next, substitution of eqref{6}-eqref{8}
into eqref{1} gives an equation in $c$
begin{align}
frac{136c^3-32c^4+256c^2+c^6-512}{c(c^2+8)^2}
&= 4
tag{9}label{9}
,
end{align}
which is equivalent to
begin{align}
(c-2)(c+4)(c^2-4c-16)(c^2-2c-4)&=0
tag{10}label{10}
end{align}
with six real roots
begin{align}
{
2,-4,
1+sqrt5, 1-sqrt5, 2+2sqrt5, 2-2sqrt5
}
tag{11}label{11}
.
end{align}
Expressions eqref{6}-eqref{8} provide corresponding values of $d,b$ and $a$.

Solving the system $b+c+d=4$, $ad+bc=-8$, $a+b=5$, $cd=-8$

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