42nd Swedish 2002 Problem 5 The reals a, b satisfy a 3 - 3a 2 + 5a - 17 = 0, b 3 - 3b 2 + 5b + 11 = 0. Find a+b.
2 Solution For x  3. But the product of the roots is 17, so the equation cannot have 3 real roots. Since it had odd degree, it has at least one real root. Non-real roots occur in complex conjugate pairs, so the equation must have just one real root. We have (2-a) 3 - 3(2-a) 2 + 5(2-a) + 11 = (8 - 12a + 6a 2 - a 3 ) + (-12 + 12a - 3a 2 ) + (10 - 5a) + 11 = 17 - 5a + 3a 2 - a 3 = 0. Hence there is a bijection between roots of the two cubics. Hence the other cubic also has just one real root and the sum of the two roots is 2. 42nd Swedish 2002 © John Scholes jscholes@kalva.demon.co.uk 21 December 2003 Last corrected/updated 21 Dec 03