The real polynomial p(x) ≡ x3 + ax2 + bx + c has three real roots α < β < γ. Show that √(a2 - 3b) < (γ - α) ≤ 2 √(a2/3 - b).
Solution
a2 - 3b = (α+β+γ)2 - (αβ + βγ + γα) = (γ - α)2 - (γ - β)(β - α) < (γ - α)2. Hence √(a2 - 3b) < (γ - α).
Also 4(a2 - 3b) - 3 (γ - α)2 = (γ - α)2 - 4(γ - β)(β - α) = ( (γ - β) + (β - α) )2 - 4(γ - β)(β - α) = ( (γ - β) - (β - α) )2 ≥ 0. Hence (γ - α) ≤ 2 √(a2/3 - b).
© John Scholes
jscholes@kalva.demon.co.uk
5 Mar 2002