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If $\ a_1,\ a_2,\ a_3,...\ a_n$ $n\ge2$are real and $(n-1)\ a_1^2 -2n\ a_2 <0$ prove that at least two roots of the equation $x^n+\ a_1x^{n-1}+\ a_2x^{n-2}+...+\ a_n=0$ are imaginary.

Manually i am able to prove it as $\ a_2$ is always positive. I tried different combination and it worked but needs logical proof from the equation.

Samar Imam Zaidi
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    Put n=2 , the discriminant is less than zero which means roots are imaginary but same stuff needs to be proved for general equation. – Samar Imam Zaidi Aug 04 '18 at 02:11
  • In the case of $n=2$, a negative discriminant only implies that the roots are non-real (that is, of the form $a+bi$, with $b\neq 0$), not that they are imaginary (that is, of the form $0+bi$). Or does the source of the question use the term "imaginary" rather imprecisely to mean "non-real"? – Blue Aug 04 '18 at 02:52

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Suppose first that all roots of the polynomial $f=x^n+ a_1x^{n-1}+ a_2x^{n-2}+\dots+ a_n$ are real and distinct. Then the derivative $f'$ has also all roots real and distinct, they are placed in the open intervals between consecutive roots. Inductively, we build the derivative till we get a polynomial of degree two. This polynomial is $$ g= f^{(n-2)}(x) = \frac {n!}{2!} \cdot x^2 + \frac {(n-1)!}{1!}\cdot a_1x + \frac {(n-2)!}{0!}\cdot a_2 \ . $$ We compute $g/(n-2)!$, which is $\frac 12n(n-1)x^2+(n-1)a_1x+a_2$, it has discriminant $<0$ by the given relation, so no real roots. Contradiction.

Our asumption is false.

The case when there are multiple roots is covered analogously, either by considering a deformation of the roots, or by observing that if $f$ has a root with algebraic multiplicity $r$, then the root survives with multiplicity $(r-1)$ in the derivative.

dan_fulea
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