Russian 2000 Problem 1 The equations x 2 + ax + 1 = 0 and x 2 + bx + c = 0 have a common real root, and the equations x 2 + x + a = 0 and x 2 + cx + b = 0 have a common real root. Find a + b + c.
-3 Solution The common root of x 2 + ax + 1 = 0 and x 2 + bx + c = 0 must also satisfy (a-b)x + (1-c) = 0, so it must be (c-1)/(a-b). Note that the other root of x 2 + ax + 1 = 0 must be (a-b)/(c-1), since the product of the roots is 1. Similarly the common root of x 2 + x + a = 0 and x 2 + cx + b = 0 must satisfy (c-1)x + (b-a) = 0, so it is x = (a-b)/(c-1). Hence x 2 + x + a = 0 and x 2 + ax + 1 = 0 have a common root. Hence it satisfies (a-1)x + (1-a) = 0. Now we cannot have a = 1, for then x 2 + ax + 1 has no real roots. Hence the common root must be 1. Hence both roots of x 2 + ax + 1 = 0 are 1 and so a = -2. So x 2 + bx + c = 0 has one root 1. Then its other root must be c/1 = c. Hence -b = 1 + c, or b + c = -1. Hence a + b + c = -3. Russian 2000 © John Scholes jscholes@kalva.demon.co.uk 20 December 2003