If the equation anxn+an−1xn−1+⋯+a1x=0 (a1=0, n≥2) has a positive root x=α, then the equation nanxn−1+(n−1)an−1xn−2+⋯+a1=0 has a positive root, which is
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Official solution
Let f(x)=anxn+⋯+a1x. f(0)=0=f(α) (given root). By Rolle's theorem, f′(x)=nanxn−1+⋯+a1=0 has a root strictly between 0 and α, i.e. smaller than α.