25th ASU 1991 Problem 18 p(x) is the cubic x 3 - 3x 2 + 5x. If h is a real root of p(x) = 1 and k is a real root of p(x) = 5, find h + k.
Put y = 2-h, where p(h) = 1, then (2-y) 3 - 3(2-y) 2 + 5(2-y) - 1 = 0, so 8-12y+6y 2 -y 3 - 12+12y-3y 2 + 10-5y - 1 = 0, or y 3 - 3y 2 + 5y = 5, or p(y) = 5. So if h is a root of p(h) = 1, then there is a root k of p(k) = 5 such that h+k = 2. To complete the proof we have to show that p(x) = 5 has only one real root. But x 3 - 3x 2 + 5x = (x-1) 3 + 2(x-1) + 3 which is a strictly increasing function of x-1 and hence of x. So p(x) = k has only one real root. Thanks to Suat Namli 25th ASU 1991 © John Scholes jscholes@kalva.demon.co.uk 2 March 2004 Last corrected/updated 2 Mar 04