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Problem: let $p(x) = x^n + a_1.x^{n-1} + a_2.x^{n-2} +......+ a_n$ where $a_i \in \mathbb R$. If $a_1 ^2 \lt a_2$, prove that not all roots of $p(x)$ can be real.

My attempt: I noticed that for $n = 2$ the claim is obviously true as the discriminant becomes less than zero. So i thought of taking an inductive approach. Define $p_k(x) = x^k + a_1.x^{k-1} +....+ a_k$.

Clearly, $p_n(x) = x.p_{n-1}(x) + a_n , \forall n\gt 2$. Now i was wondering, if a polynomial has imaginary roots, does shifting it by adding a constant lead to producing real roots? Or does there always remain imaginary roots on shifting?

Lelouch
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    Induction comes to mind, but my experience is that it rarely works with problems like this. Relating roots of $p_n$ to those of $p_{n-1}$ is too messy. One might also try factoring out one of the roots (assuming it is real), but then the coefficients $a_1$ and $a_2$ change in a way that may be difficult to manage. Using quadratics as a test case is a clue. For the roots to be real we need $a_1^2>4a_2$. So there may actually be a some slack. – Jyrki Lahtonen Apr 14 '17 at 05:18

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If the zeros are $r_1,r_2,\ldots,r_n$, then $$ a_1=r_1+r_2+\cdots+r_n $$ and $$ a_2=r_1r_2+\cdots+r_{n-1}r_n=\sum_{1\le i<j\le n}r_ir_j. $$ Therefore $$ r_1^2+r_2^2+\cdots+r_n^2=a_1^2-2a_2. $$ But $a_1^2<a_2$ implies that $a_1^2-2a_2<0$.

Jyrki Lahtonen
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