## Task A-2.7.

The quadratic equation $x^{2}+b x+c=0$ has real solutions whose sum of squares is equal to 2. Determine all possible values of the expression $b+c$.
## Solution.

According to Viète's formulas, for the solutions $x_{1}, x_{2}$ of the given quadratic equation, we have

$$
x_{1}^{2}+x_{2}^{2}=\left(x_{1}+x_{2}\right)^{2}-2 x_{1} x_{2}=b^{2}-2 c
$$

Thus, we have $b^{2}-2 c=2$, which means $c=\frac{b^{2}-2}{2}$.

Notice that

$$
b+c=b+\frac{b^{2}-2}{2}=\frac{b^{2}+2 b-2}{2}=\frac{(b+1)^{2}-3}{2}
$$

The given equation has real solutions, so it is necessary that $b^{2}-4 c \geqslant 0$.

Since $c=\frac{b^{2}-2}{2}$, the condition $b^{2}-4 c \geqslant 0$ becomes $b^{2} \leqslant 4$, or $-2 \leqslant b \leqslant 2$.

From the condition $-2 \leqslant b \leqslant 2$, we sequentially get

$$
\begin{gathered}
-1 \leqslant b+1 \leqslant 3 \\
0 \leqslant(b+1)^{2} \leqslant 9 \\
-3 \leqslant(b+1)^{2}-3 \leqslant 6 \\
-\frac{3}{2} \leqslant \frac{(b+1)^{2}-3}{2} \leqslant 3
\end{gathered}
$$

Since $b+c=\frac{(b+1)^{2}-3}{2}$, we conclude that $b+c$ can take any value $u$ in the interval $\left[-\frac{3}{2}, 3\right]$.

## SCHOOL/CITY COMPETITION IN MATHEMATICS

3rd grade - high school - A variant
February 4, 2022.

## IF A STUDENT HAS A DIFFERENT APPROACH TO SOLVING THE PROBLEM, THE COMMITTEE IS OBLIGED TO GRADE AND EVALUATE THAT APPROACH IN AN APPROPRIATE MANNER.