Russian 2001

x³ + ax² + bx + c has three distinct real roots, but (x² + x + 2001)³ + a(x² + x + 2001)² + b(x² + x + 2001) + c has no real roots. Show that 2001³ + a 2001² + b 2001 + c > 1/64.

Solution

We can write x² + x + 2001 = (x + 1/2)² + 2000 - 1/4, so it takes any value ≥ 2000 - 1/4. Hence the roots of x³ + ax² + bx + c must all be < 2000 - 1/4. Suppose the roots are p, q, r. We have p, q, r < 2000 - 1/4, hence 2000-p > 1/4. Similarly, 2000-q > 1/4 and 2000-r > 1/4. Multiplying we get 2001³ + a 2001² + b 2001 + c > 1/64.

Thanks to Suat Namli

Russian 2001

© John Scholes
jscholes@kalva.demon.co.uk
7 February 2004
Last corrected/updated 7 Feb 04