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Consider the polynomial equation aₙ xⁿ + aₙ₋₁xⁿ⁻¹ + ⋯ + a₁ x = 0, where a₁ ≠ 0 and n ≥ 2. If it has a positive root x = α, then the related equation n aₙ xⁿ⁻¹ + (n-1)aₙ₋₁xⁿ⁻² + ⋯ + a₁ = 0 has a positive root that is

1. greater than α
2. less than α
3. greater than or equal to α
4. equal to α

Correct answer: less than α

Solution

The polynomial p(x) has no constant term, so p(0)=0, and it also has p(alpha)=0. The second equation is p'(x)=0. By Rolle's theorem applied on [0, alpha], p' has a root strictly between 0 and alpha, i.e. a positive root less than alpha.

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