# Task 3. (12 points)

Find all pairs $(a, b)$ of real numbers $a$ and $b$ such that the equations $x^{2}+a x+b^{2}=0$ and $x^{2}+b x+a^{2}=0$ have at least one common root.

#
# Solution.

Let $x_{0}$ be the common root of the equations.

Substituting $x_{0}$ into the equations and subtracting one from the other, we get

$$
(a-b) x_{0}+b^{2}-a^{2}=0 \Leftrightarrow(a-b)\left(x_{0}-(a+b)\right)=0
$$

1) Suppose $a \neq b$, then $x_{0}=a+b$ and

$(a+b)^{2}+a(a+b)+b^{2}=0$ or $2 a^{2}+3 a b+2 b^{2}=0$ or $2\left(a+\frac{3}{4} b\right)^{2}+\frac{7}{8} b^{2}=0$.

From which it follows that $a=b=0$, a contradiction.

2) Suppose $a=b$, then both equations have the form $x^{2}+a x+a^{2}=0$ with discriminant $-3 a^{2} \leq 0$. Therefore, the only possibility is $a=b=0$.

Answer. $(0,0)$

| Criterion | Evaluation | Points |
| :---: | :---: | :---: |
| Complete solution. | + | 12 |
| The main idea of the solution is found. It is shown that there are no other solutions. The solution is incomplete or contains deficiencies. The answer is correct. | $\pm$ | 9 |
| The main idea of the solution is found. The answer is correct. It is not shown that there are no other solutions. | $+/ 2$ | 6 |
| The correct answer is given. The solution is missing or does not contain progress in the right direction. | $\mp$ | 2 |
| The solution does not meet any of the criteria described above. | $-/ 0$ | 0 |
| Maximum score |  | 12 |