Prove that the largest number of real roots of the equation $ x^5+a_1x^4+a_2x^3+a_3x^2+a_4x+a_5=0$ whose coefficients are real,is three if $2a_1^2-5a_2<0.$
My attempt is:
As coefficients are real,so complex roots will come in pair.Either one or three or five real roots are possible.I does not know what is relation between number of real roots and coefficients.Can someone guide me?I am cofused.
Best Answer
Assuming that all the roots $\xi_1,\xi_2,\xi_3,\xi_4,\xi_5$ of your polynomial are real, $$ (\xi_1+\xi_2+\xi_3+\xi_4+\xi_5)^2 \leq 5(\xi_1^2+\xi_2^2+\xi_3^2+\xi_4^2+\xi_5^2) \tag{1}$$ must hold: it is the Cauchy-Schwarz inequality. On the other hand, by Viète's theorem the LHS of $(1)$ is $a_1^2$, while the RHS of $(1)$ is $5(a_1^2-2a_2)$. So if all the roots of the polynomial are real, $$ 2a_1^2 \geq 5a_2 \tag{2}$$ must hold. We have that $(2)$ does not hold, hence the real roots of our quintic polynomial are at most $3$ (since complex roots come in conjugated pairs).