Under what conditions on cofficients of a polynomial $p(x)$, the roots of $p(x)$ are real and positive?
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Discriminant is sufficient i suppose – Archis Welankar Apr 18 '16 at 17:34
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One can also use general theory of equations such that we get only real roots – Archis Welankar Apr 18 '16 at 17:35
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@ArchisWelankar: How do you define discriminant for a polynomial of $n$th degree? – Ángel Mario Gallegos Apr 18 '16 at 17:35
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1https://en.wikipedia.org/wiki/Sturm%27s_theorem#Number_of_real_roots – Erick Wong Apr 18 '16 at 17:38
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You also have the very nice Sylvester-Hermite theorem:
Let $P(X)\in \mathbf R[X]$ a separable polynomial of degree $n$. Let $\zeta_1,\dots,\zeta_n$ be the (real or complex) roots of $P(X)$. Consider the Newton's sums $p_k=\zeta_1^k+\dots,\zeta_n^k,\quad k\ge 0$. To the real symmetric matrix $$M=\begin{bmatrix} p_0&p_1&\dots& p_{n-1}\\ p_1&p_2&\dots& p_n\\[-1ex] \vdots&\vdots&&\vdots\\[-1ex] p_{n-1}& p_n&\dots& p_{2n-2} \end{bmatrix}$$ is associated a quadratic form on $\mathbf R^n$.If $(s,t)$ is the signature of this quadratic form, the number of real roots of $P(X)$ is $\;s-t$.
Bernard
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