1. Given the equations

$$
x^{2}-a x+b-4=0 \text { and } y^{2}-b y+a-\frac{1}{4}=0, a, b \in \mathbb{R}
$$

For which values of $a$ and $b$, the roots of the second equation are the reciprocal values of the roots of the first equation?
Solution. Let $x>0$ and $\beta$ be roots of the first equation. According to the conditions of the problem, $\alpha, \beta \neq 0$. From Vieta's formulas, for the first equation we have: $\alpha+\beta=a$ and $\alpha \beta=b-4$, and for the second: $\frac{1}{\alpha}+\frac{1}{\beta}=b$ and $\frac{1}{\alpha} \cdot \frac{1}{\beta}=a-\frac{1}{4}$. Then, by substituting the first two equations into the last two, we get: $a=b(b-4)$ and $1=(b-4)\left(a-\frac{1}{4}\right)$. From the last two equations, we obtain $(b-4)\left(b(b-4)-\frac{1}{4}\right)=1$, which gives us the equation $b^{3}-8 b^{2}+\frac{63}{4} b=0$. Its solutions are: $b_{1}=0, b_{2}=\frac{7}{2}, b_{3}=\frac{9}{2}$ and the corresponding values for $a$ are $a_{1}=0, a_{2}=-\frac{7}{4}, a_{3}=\frac{9}{4}$.