How many real roots does 2x = 1 + x2 have?
Answer: 3. x = 0, x = 1 and a value just over 4.
Clearly there are no roots for negative x, since for such x, 2x < 1, whereas 1 + x2 > 1. There are certainly roots at x = 0 and 1. Also 24 < 42 + 1, whereas 25 > 52 + 1, so there is a root between 4 and 5. We have to show that there are no other roots. Put f(x) = 2x - x2 - 1. Then f ''(x) = (ln 2)2 2x - 2. This is strictly increasing with a single zero. f '(0) > 0, so f '(x) starts positive, decreases through zero to a minimum, then increases through zero. So it has just two zeros. Hence f(x) has at most three zeros, which we have already found.
34th Putnam 1973
© John Scholes
22 Aug 2001