If the equation $a_n{x}^n+a_{n-1}{x}^{n-1}+\cdots +a_{1}x=0,a_1\ne 0,n\ge 2$, has a positive root $x=\alpha$, then the equation $n{a}_n{x}^{n-1}+(n-1)a_{n-1}{x}^{n-2}+\cdots +a_1=0$ has a positive root, which is:

Let $f(x)=a_n{x}^n+a_{n-1}{x}^{n-1}+\cdots +a_{1}x$.The equation $f(x)=0$ has roots $x=0$ and $x=\alpha$.Since $f(x)$ is a polynomial, it is continuous and differentiable everywhere.By Rolle's Theorem, if $f(0)=f(\alpha )=0$, then there exists at least one $c\in (0,\alpha )$ such that $f^{\prime }(c)=0$.$f^{\prime }(x)=n{a}_n{x}^{n-1}+(n-1)a_{n-1}{x}^{n-2}+\cdots +a_1$.Thus, $f^{\prime }(x)=0$ has a positive root $c$ which is less than $\alpha$.