Tell me more ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.
up vote 1 down vote favorite

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?

share|improve this question
2  
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
1  
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
1  
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

3 Answers

up vote 1 down vote

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.

share|improve this answer
up vote 1 down vote

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.

share|improve this answer
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
show 2 more comments
up vote 0 down vote

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

$x_1*x_2=b$

$x_2+x_2=a$

share|improve this answer

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.