Find all the possible real values for $a,b,c,d$.

Question

Let pairs $(a,c)$ and $(b,d)$ be roots of the equations $x^2 + ax - b = 0$ and $x^2 + cx + d = 0$ respectively. Find all possible real values for $a,b,c,d$.

Answer 1

By Vieta's formulas we have $$ a + c = -a\\ ac = -b\\ b + d = -c\\ bd = d $$ From last equation, $b = 1$ or $d = 0$.

I) $b=1$. So, $$ a + c = -a\\ ac = -1\\ 1 + d = -c $$ From second eq. $c = -1/a$ and $$ a - \frac1a = -a \Longrightarrow 2a^2 = 1\\ d = \frac1a - 1 $$ So, $a = 1/\sqrt2, b = 1, c =- \sqrt2, d = \sqrt2-1$ and $a = -1/\sqrt2, b = 1, c = \sqrt2, d = -\sqrt2-1$ are solutions.

II) $d=0$. So, $$ a + c = -a\\ ac = -b\\ b = -c $$ or $$ a + c = -a\\ ac = c\\ b = -c $$ From second eq. $c=0$ or $a=1$.

II.1) $c=0$; we have $$ a = -a\\ b = 0 $$ and solution is $a=0, b = 0, c = 0, d =0$.

II.2) $a=1$; we have $$ 1 + c = -1,\\ b = -c, $$ and $a=1, b=2, c=-2, d=0$ is solution.

Answer 2

The sum of the roots of the first equation is $$a+c=-a$$ so $c=-2a$. And the product of the roots of the second equation is $$bd=d$$ so $d=0$ or $b=1$.

If $d=0$ the second equation is $x^2+cx$ whose roots are $0$ and $-c$. That is, $b=0$ or $b=-c$. If $b=0$ the roots of the first equation are $0$ and $-a$, so $a=0$ and $c=0$. This gives $$a=b=c=d=0$$ If $b=-c$ then $ac=c$, that is, $-2a^2=-2a$. For $a=0$ we get the same solution, and for $a=1$, $$a=1,b=2,c=-2,d=0$$

If $b=1$ The equations are $x^2+ax-1=0$ and $x^2-2ax+d=0$. Then $b+d=2a$, or $d=2a-1$. Also, $ac=-1$ and therefore $2a^2=1$. This give us two more solutions: $$a=\frac1{\sqrt 2},b=1,c=-\sqrt 2,d=\sqrt 2-1$$ $$a=-\frac1{\sqrt 2},b=1,c=\sqrt 2,d=-\sqrt 2-1$$