IMO 1979 SL 6

Origin: CZS

Problem

Find the real values of p for which the equation 2p + 1 −x2 + 3x + p + 4 = x2 + 9x + 3p + 9 in x has exactly two real distinct roots ( \sqrt t means the positive square root of t).

Solution

Setting q = x2 + x −p, the given equation becomes

(x + 1)2 −2q +

(x + 2)2 −q =

(2x + 3)2 −3q. (1) Taking squares of both sides we get 2

((x + 1)2 −2q)((x + 2)2 −q) = 2(x + 1)(x + 2). Taking squares again we get q
2q −2(x + 2)2 −(x + 1)2 = 0. If 2q = 2(x + 2)2 + (x + 1)2, at least one of the expressions under the three square roots in (1) is negative, and in that case the square root is not well-defined. Thus, we must have q = 0. Now (1) is equivalent to |x + 1| + |x + 2| = |2x + 3|, which holds if and only if x ̸\in(−2, −1). The number of real solutions x of q = x2 + x −p = 0 which are not in the interval (−2, −1) is zero if p < −1/4, one if p = −1/4 or 0 < p < 2, and two otherwise. Hence, the answer is −1/4 < p \leq0 or p \geq2.