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)
 
Here is a little puzzle from the book 100 Geometric Games by Pierre Berloquin. The side of a small square is one meter long and the side of a larger square one and a half meters long. One vertex of the large square is at the center of the small square. The side of the large square cuts two sides of the small square into one- third parts and two-thirds parts. What is the area where the squares overlap?

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