69th Belarusian Mathematical Olympiad

Question

The polynomial f(x) = x^2 + bx + 8, b 0, has two different real roots x_1 and x_2. The numbers x_1 + 1/x_2 and x_2 + 1/x_1 are the roots of the monic polynomial g(x) of degree 2. Find all possible values of g(1) if g(1) = f(1).

Accepted Answer

Answer: -8. Vieta's formulas for f(x) imply x_1 + x_2 = -b and x_1 x_2 = 8. Let g(x) = x^2 + cx + d, then from Vieta's formulas for g(x): aligned -c &= x_1 + 1x_2 + x_2 + 1x_1 = -b + x_1 + x_2x_1 x_2 = -b - b8 d &= (x_1 + 1x_2) (x_2 + 1x_1) = x_1 x_2 + 1x_1 x_2 + 1 + 1 = 8 + 18 + 2 = 10 + 18 aligned The condition g(1) = f(1) is equivalent to 1 + b + b8 + 10 + 18 = 1 + b + 8, whence 178 + b8 = 0, i.e. b = -17. Therefore, g(1) = f(1) = 1 + b + 8 = 1 - 17 + 8 = -8.