Consider the biquadratic polynomial equation $\rho_0y^4+\rho_1y^3+\rho_2y^2+\rho_3y+\rho_4y=0$, where $\rho_0,\rho_1,\rho_2, \rho_4$ are positive and $\rho_3$ is negative. So by Descartes' rule of signs it has either two positive roots or no positive root.
Now what is the necessary condition (on the coefficient $\rho_i, i=0,1,..4$ ) for which the above biquadratic equation must have two positive roots??
Best Answer
Let $f(z)=ax^4+bx^3+cx^2+dx+e$ be a quartic polynomial with real coefficients and $a>0$, $e>0$.
Define:
$\alpha = ba^{-3/4}e^{-1/4}$
$\beta=ca^{-1/2}e^{-1/2}$
$\gamma=da^{-1/4}e^{-3/4}$
Then $f(x)\geq0$ for all $x>0$ the following conditions are sufficient for positivity:
Hope this help you.
Regards
Ric