Two Quadratics, the roots of one are the coefficients of the other and vice versa.... can they exist?

Question

I spent hours on this... feel pretty pathetic. I read this on an old AMATYC exam:

Let $\quad c, \quad$ and $\quad d\quad$ be the roots of $\quad x^2+ax+b$

and let $\quad a, \quad$ and $\quad b\quad$ be the roots of $\quad x^2+cx+d$

What is the value of $$a+b+c+d$$ Somehow, I am to reason that this value is $\quad -2$

I do realize that this must mean that $$\begin{cases} a&=-(c+d) \\ b&=cd \\ c&=-(a+b) \\ d&=ab \\ \end{cases}$$ would have a solution, but I cannot show one exists, can you find an example?...

the first and third equations alone imply that you can set two of the four variables free: $$\begin{pmatrix}1&0&1&1 \\ 1&1&1&0\end{pmatrix}\cdot \begin{pmatrix}a\\b\\c\\d\end{pmatrix}=0 \quad \sim \quad\begin{pmatrix}1&0&1&1 \\ 0&1&0&-1\end{pmatrix}\cdot \begin{pmatrix}a\\b\\c\\d\end{pmatrix}=0$$ so that $$\begin{cases} a&=-c-d \\ b&=d \\ c&=c \\ d&=d\\ \end{cases}$$

perhaps new letters?... kidding but:

$$x^2-(r_1+r_2)x+r_1r_2 \qquad \text{vs}\qquad x^2+(r_1+r_2-r_1r_2)x-r_1r_2(r_1+r_2)$$

works only one way: the roots of second are the coefficients of the first alright, but not the other way around.

Answer 1

Picking up where the OP left off.

$$\begin{cases}\begin{align} a&=-(c+d) \quad\quad\quad\quad\tag{1}\\ b&=cd \quad\quad\quad\quad\tag{2} \\ c&=-(a+b) \quad\quad\quad\quad\tag{3}\\ d&=ab \quad\quad\quad\quad\tag{4}\\ \end{align}\end{cases}$$

Subtracting $\,(1)-(3)\,$:

$$ \require{cancel} \cancel{a} - \bcancel{c} = -(\bcancel{c}+d) + (\cancel{a}+b) \quad\iff\quad b = d $$

Using $\,b = d\,$ in $\,(2)\,$, $\,(4)\,$:

$$ b = cb \quad\iff\quad b = 0 \;\;\lor\;\; c = 1 \\ b = ab \quad\iff\quad b = 0 \;\;\lor\;\; a = 1 $$

• $b = d = 0 \implies c = -a\,$ from $(1)$ and therefore $a+b+c+d = 0\,$.

• $b = d \ne 0 \implies a = c = 1 \implies a + b + c + d = -(a+c) = -2$ from $\,(1) + (3)\,$.

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[ EDIT ] $\;$ For a brute force approach using heavier machinery, let $\,t=a+b+c+d\,$ be the sum in question. Eliminating $a,b,c,d$ between $(1)-(4)$ and the expression for $t$ results (courtesy WA) in the equation $\,t(t+2)= 0\,$ with roots $0, -2$.

Answer 2

So $a^2+ca+d=0$ and substituting $a=-(c+d)$ we get $(c+d)^2-c(c+d)+d=0 \implies d(c+d+1)=0 \implies c+d=-1$.
Note that if $d=0$ we have that $a^2+ca=0 \implies c=a=b=d=0$. Similarly, $a+b=-1$

Answer 3

$a+c+d = a+b+c = 0$ implies $b=d=-(a+c)$. If $b=d\neq 0$, then $a=c=1$ and $a+b+c+d = -2$. However, if $b=d=0$, then $c=-a$ works and $a+b+c+d = 0$.

The pairs of equations are $(x^2+x-2=0, x^2+x-2=0)$ with roots $1$ and $-2$, and $(x^2+ ax+ 0=0, x^2-ax+0=0)$: the first has roots $-a$ and 0, and the second has roots $a$ and 0.