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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?

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Please consider accepting some of the answers to your previous questions as a thank you to the ones who helped you. – Gigili Jul 29 '12 at 14:34
I've edited your question to more clearly express what I think you mean. If I interpreted your question incorrectly, please edit it again to clarify what you meant. – Zev Chonoles Jul 29 '12 at 14:34
People who downvote: please provide feedback to OP. S/he might be oblivious to why his/her question needs improvement. – user2468 Jul 30 '12 at 23:05
@J.D. OP.S?Sorry ,I don't understand you. – tan9p Jul 31 '12 at 11:33
@tan9p OP: means "original poster." My comment is directed to users who downvoted your question without leaving feedback. – user2468 Jul 31 '12 at 15:45

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.

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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.

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I would have thought that the wrapping around was the hard part :-) – joriki Jul 31 '12 at 5:13
@J.D. but ${a_1,b_1}$ may not the roots of one equation. – tan9p Jul 31 '12 at 11:29
@joriki You're right. Is it possible to do a fixed-point iteration scheme? In the 1st iteration, take roots of $p_1$ to be random & construct the $n$ polynomials. Then in the 2nd iteration take roots of $p_1$ to be $a_n, b_n.$ Propagate through & repeat the construction until it converges to a set of $a_i$ and $b_i$'s such that $a_n, b_n$ are roots of $p_1$...?! – user2468 Jul 31 '12 at 15:51
@tan9p Do you mean $a_n, b_n$? In my answer $a_1, b_1$ are roots of $p_2,$ but $a_n, b_n$ are not roots of any polynomial (yet). We desired them to be roots of $p_1.$ – user2468 Jul 31 '12 at 15:53
@J.D. I mean ,er, to consider this situation:$a_1,b_2$ are the roots of $x^2-a_3x+b_3=0$,$a_2,b_1$ are the roots of $x^2-a_4x+b_4=0$,and $a_3,b_3$ are the roots of $x^2-a_1x+b_1=0$,$a_4,b_4$ are the roots of $x^2-a_2x+b_2=0$. – tan9p Jul 31 '12 at 16:15

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



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