Given a system of quadratic equations $x^2-a_ix+b_i=0$, can all of the coefficients $a_i$, $b_j$ be solution to  one of  these above equation? Here are $n$ quadratic equations ($n>1$): $$x^2-a_ix+b_i=0\quad(i=1,\ldots, n)$$ where the $a_i$, $b_i$ are distinct. Can all of the $a_i,b_i$ be roots of one of the above equations?
 A: In general the answer is NO: Assume this is possible, then $\{a_i , i=1..n\}=\{a_i+b_i, i=1..n\}$ and $\{b_i, i=1..n\}=\{a_i*b_i, i=1..n\}$, and we obtain as necessarily conditions $\sum {b_i} = 0$ and $\prod{a_i}=1$. Can't say more for the moment.
A: I don't fully understand the question, but I will try:
Recall a polynomial of degree $2$ over $\Bbb C$ has two roots. If we are allowed to construct the system of $n$ of polynomials, then we can force the desired root properties as follows.
Let's start with an arbitrary polynomial call it $p_1 = x^2 - a_1 x + b_1.$ Now to construct $p_2 = x^2 - a_2 x + b_2$ such that $\{a_1, b_1 \}$ are roots of $p_2,$ we need to have $$ a_2 = a_1 + b_1, b_2 = a_1 b_1. \tag{1}$$
And so forth, the polynomial $p_i$ will have roots $\{a_{i-1}, b_{i-1} \}.$
We need to do something to wrap around $p_n$ and $p_1.$ But I will leave that to you. Does this help?

Update: if all you want is sufficient conditions on $a_i$'s and $b_i$'s such that all $a_i, b_i$ are roots of some $p_j.$ Then we can generalize $(1)$ into the following $n$ conditions; for a given system of polynomials $p_i = x^2 - a_i x + b_i, 1 \le i \le n,$ the following have to hold:
$$ a_1 = a_n + b_n, \quad b_1 = a_n b_n \\
   a_2 = a_1 + b_1, \quad  b_2 = a_1 b_1 \\
   a_3 = a_2 + b_2, \quad  b_3 = a_2 b_2 \\ 
   \ldots \\
   a_{n} = a_{n-1} + b_{n-1}, \quad b_n = a_{n-1} b_{n-1} $$
The explicit construction I gave doesn't satisfy the first condition though.
A: so if we are talking about complex numbers or roots could be complex numbers,then yes,otherwise for real solution discriminant
$D=a^2-4*b$ must be nonnegative and also using vietas theorem,roots could be satisfies
$x_1*x_2=b$
$x_2+x_2=a$
