## Task A-2.5.

Determine all pairs $\{a, b\}$ of distinct real numbers such that the equations

$$
x^{2} + a x + b = 0 \quad \text{and} \quad x^{2} + b x + a = 0
$$

have at least one common solution in the set of real numbers.
## First Solution.

Let the common solution be $\mathrm{s} x_{0}$. By equating the equations, we get:

$$
\begin{aligned}
x_{0}^{2}+a x_{0}+b & =x_{0}^{2}+b x_{0}+a \\
x_{0} a-x_{0} b+b-a & =0 \\
x_{0}(a-b)-(a-b) & =0 \\
\left(x_{0}-1\right)(a-b) & =0
\end{aligned}
$$

From the condition $a \neq b$, we get $x_{0}=1$.

Substituting into the equations, we obtain the necessary condition for $a$ and $b$:

$$
1+a+b=0
$$

Thus, the number 1 is a solution to the quadratic equations if and only if the above relation holds. 1 point

Among such pairs, we need to exclude those for which $a=b=-1-a$, i.e., $a=-\frac{1}{2}$.

Therefore, the solutions are all pairs $\{a, b\}$ such that $a+b=-1$, except when $a=b=-\frac{1}{2}$.