Let f(x)=anxn+an−1xn−1+......+a1x=0 The other given equation, nanxn−1+(n−1)an−1xn−2+.........+a1=0=f′(x) Given a1≠0⇒f(0)=0 Againf(x) has rootα,⇒f(α)=0 ∴f(0)=f(α) ∴ By Rolle's theorem f′(x)=0 has root between (0,α) Hence f′(x) has a positive root smaller than α.