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

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The equation presented is a polynomial with real coefficients, specifically structured as \(x^n + a_1x^{n-1} + a_2x^{n-2} + \ldots + a_n = 0\), where it is established that \(a_1^2 < a_2\). By applying Newton's identities, the sum of the squares of the roots, \(x_1^2 + x_2^2 + \ldots + x_n^2\), is equal to \(a_1^2 - a_2\). Since \(a_1^2 < a_2\), this sum is negative, indicating that not all roots can be real, as the sum of squares of real numbers must be non-negative. The conclusion drawn is that the condition \(a_1^2 < a_2\) ensures the existence of non-real roots in the polynomial.
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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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