Competition problem — Algebra — Maths Olympiad Prep

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Algebra Difficulty 5.8 AIME, harder Prove it Belarus

The polynomial $f(x) = x^2 + bx + 8$ , $b \neq 0$ , 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 $g(1) = f(1)$ .

Solution

Answer: $-8$ .

Vieta's formulas for $f(x)$ imply $x_1 + x_2 = -b$ and $x_1 \cdot x_2 = 8$ . Let $g(x) = x^2 + cx + d$ , then from Vieta's formulas for $g(x)$ :
$\begin{aligned} -c &= x_1 + \frac{1}{x_2} + x_2 + \frac{1}{x_1} = -b + \frac{x_1 + x_2}{x_1 \cdot x_2} = -b - \frac{b}{8} \\ d &= \left(x_1 + \frac{1}{x_2}\right) \cdot \left(x_2 + \frac{1}{x_1}\right) = x_1 x_2 + \frac{1}{x_1 x_2} + 1 + 1 = 8 + \frac{1}{8} + 2 = 10 + \frac{1}{8} \end{aligned}$
The condition $g(1) = f(1)$ is equivalent to $1 + b + \frac{b}{8} + 10 + \frac{1}{8} = 1 + b + 8$ , whence $\frac{17}{8} + \frac{b}{8} = 0$ , i.e. $b = -17$ . Therefore, $g(1) = f(1) = 1 + b + 8 = 1 - 17 + 8 = -8$ .

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