## Task A-2.2.

The equation

$$
\left(x^{2}+a x+b\right)\left(x^{2}+a x+c\right)=0
$$

has four distinct real solutions, which are $a, b, c$ and $-1$. Determine the numbers $a, b$, and $c$.
## First Solution.

Two of the four numbers $a, b, c \text{ and } -1$ are solutions to the quadratic equation $x^{2} + a x + b = 0$, while the other two are solutions to the quadratic equation $x^{2} + a x + c = 0$.

The sum of the solutions of these equations is the sum of the solutions of each of the quadratic equations $x^{2} + a x + b = 0$ and $x^{2} + a x + c = 0$. According to Vieta's formulas, this sum is $-2a$. Since $a, b, c$ and $-1$ are all four solutions of these equations, we conclude

$$
-2a = a + b + c - 1
$$

1 point

Similarly, the product of these four solutions is the product of the constant terms $b$ and $c$, but it is also equal to the product of all four numbers $a, b, c$ and $-1$. Therefore, we have

$$
bc = -abc
$$

1 point

From the last equation, it follows that $(a + 1)bc = 0$. The case $a = -1$ is discarded because, according to the problem's condition, all four solutions must be distinct. We conclude that one of the numbers $b$ and $c$ must be zero.

Notice that if the triplet $(a, b, c)$ satisfies the conditions of the problem, then the triplet $(a, c, b)$ also satisfies the conditions of the problem. Therefore, without loss of generality, we can assume that $c = 0$.

From the equation for the sum of the solutions, we get

$$
\begin{aligned}
-2a & = a + b - 1 \\
b & = -3a + 1
\end{aligned}
$$

The solutions to the quadratic equation $x^{2} + a x + c = x(x + a)$ are $c = 0$ and the number $-a$. The second solution cannot be equal to $a$, because then $a$ would be zero. Therefore, $a$ is a solution to the quadratic equation $x^{2} + a x + b = 0$.

Substituting the number $a$ into this equation, we get

$$
0 = a^{2} + a \cdot a + b = 2a^{2} - 3a + 1
$$

The solutions to this equation are $a = 1$ and $a = \frac{1}{2}$.

Using the equation $b = -3a + 1$, in the first case we get $b = -2$, and in the second case we get $b = -\frac{1}{2}$.

By direct verification, we see that both possibilities for the numbers $a, b$, and $c$ indeed satisfy the conditions of the problem. Namely, the solutions to the equation

$$
\left(x^{2} + x - 2\right)\left(x^{2} + x + 0\right) = 0
$$

are the numbers $a = 1$ and $b = -2$ (roots of the quadratic expression in the first parenthesis), and $c = 0$ and $-1$ (roots of the quadratic expression in the second parenthesis), while the solutions to the equation

$$
\left(x^{2} + \frac{1}{2} x - \frac{1}{2}\right)\left(x^{2} + \frac{1}{2} x + 0\right) = 0
$$

are the numbers $a = \frac{1}{2}$ and $-1$ (roots of the quadratic expression in the first parenthesis), and $c = 0$ and $b = -\frac{1}{2}$ (roots of the quadratic expression in the second parenthesis).

We conclude that all the desired triplets of numbers $(a, b, c)$ are

$$
(1, -2, 0), (1, 0, -2), \left(\frac{1}{2}, -\frac{1}{2}, 0\right), \left(\frac{1}{2}, 0, -\frac{1}{2}\right)
$$