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.

Mathematics Asked by Karagum on December 28, 2020

2 AnswersFrom 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.

Correct answer by Somos on December 28, 2020

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 $a$.

Answered by g.kov on December 28, 2020

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