1. Determine all pairs $(a, b)$ of real numbers for which each of the equations

$$
x^{2}+a x+b=0, \quad x^{2}+(2 a+1) x+2 b+1=0
$$

has two distinct real roots, and the roots of the second equation are the reciprocals of the roots of the first equation.
SOLUTION. Let $x_{1}, x_{2}$ be the roots of the first equation. Then

$$
x_{1}+x_{2}=-a, \quad x_{1} x_{2}=b
$$

and since the second equation has roots $1 / x_{1}$ and $1 / x_{2}$, we have

$$
\frac{1}{x_{1}}+\frac{1}{x_{2}}=-(2 a+1), \quad \frac{1}{x_{1}} \cdot \frac{1}{x_{2}}=2 b+1 .
$$

Thus, $\frac{1}{b}=2 b+1$, which leads to the quadratic equation $2 b^{2}+b-1=0$, with roots $b=-1$ and $b=\frac{1}{2}$.

For $b=-1$, we have

$$
-(2 a+1)=\frac{1}{x_{1}}+\frac{1}{x_{2}}=\frac{x_{1}+x_{2}}{x_{1} x_{2}}=\frac{-a}{-1}
$$

which is a linear equation in $a$ with the solution $a=-\frac{1}{3}$.

Similarly, for $b=\frac{1}{2}$, we get $-(2 a+1)=-2 a$, but this equation has no solution. By verification (it is necessary to check that the roots are real), we confirm that the pair $a=-\frac{1}{3}$, $b=-1$ is the (only) solution to the problem.

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