50th Mathematical Olympiad in Ukraine, Fourth Round (March 23, 2010)

Question

Let P(x), Q(x) and R(x) be polynomials, such that Q(x) and R(x) take nonnegative values only. It is known, that the equation P(x) + Q(x) + Q(x) + R(x) = 0 has infinitely many solutions. Is it true, that every real number is a root of this equation?

Accepted Answer

Not necessary.

We construct the following example:

P (x) = x

,

Q (x) = \frac{1}{4} x^{2}

,

Q (x) = \frac{1}{2} ∣ x ∣

,

R (x) = 0

. Then,

Q (x) + R (x) = \frac{1}{2} ∣ x ∣

and we have:

P (x) + Q (x) + Q (x) + R (x) = x + ∣ x ∣ = {2 x, 0, x \geq 0, x < 0.

Hence, our left hand side does not equal to

0

for all

x

, but the equation has infinitely many solutions.

Note, that this example is not unique, one can also take

Q (x) = \frac{1}{8} x^{2}

,

R (x) = (\frac{1}{8} x^{2})^{2}

.