Not all the roots are real iff a^2_1<a_2

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The discussion centers on the polynomial equation $$x^n + a_1x^{n-1}+a_2x^{n-2}+…+a_n = 0$$ with real coefficients, specifically under the condition that $a_1^2 < a_2$. It is established that not all roots of the polynomial can be real due to the application of Newton's identities, which show that the sum of the squares of the roots, represented as $x_1^2 + x_2^2 + \ldots + x_n^2$, equals $a_1^2 - a_2$. Since $a_1^2 < a_2$ results in a negative sum, it conclusively proves that at least one root must be non-real.

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Given the equation
$$x^n + a_1x^{n-1}+a_2x^{n-2}+…+a_n = 0$$
- with real coefficients, and $a_1^2 < a_2$.

Show that not all the roots are real.
 
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lfdahl said:
Given the equation
$$x^n + a_1x^{n-1}+a_2x^{n-2}+…+a_n = 0$$
- with real coefficients, and $a_1^2 < a_2$.

Show that not all the roots are real.
[sp]If the roots are $x_1,x_2,\ldots,x_n$ then by one of Newton's identities $x_1^2 + x_2^2 + \ldots + x_n^2 = a_1^2 - a_2$. If all the roots are real then the sum of their squares would have to be positive. But if $a_1^2 < a_2$ then that sum is negative. So not all the roots can be real.

[/sp]
 
Opalg said:
[sp]If the roots are $x_1,x_2,\ldots,x_n$ then by one of Newton's identities $x_1^2 + x_2^2 + \ldots + x_n^2 = a_1^2 - a_2$. If all the roots are real then the sum of their squares would have to be positive. But if $a_1^2 < a_2$ then that sum is negative. So not all the roots can be real.

[/sp]

Thankyou, Opalg, for your participation. Your solution is - of course - correct.(Cool)
 

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