## Task B-2.1.

Determine the quadratic function with rational coefficients that has a minimum value of -4, and one of its roots is $\sqrt{2}-1$.
## First Solution.

Let's write the quadratic function in the form $f(x)=a x^{2}+b x+c, a, b, c \in \mathbb{Q}, a \neq 0$.

If $x_{1}=\sqrt{2}-1$ is one of its roots, the other root is $x_{2}=-\sqrt{2}-1$.

By applying Viète's formulas, we get

$$
\begin{gathered}
x_{1}+x_{2}=-\frac{b}{a}=-2 \\
x_{1} x_{2}=\frac{c}{a}=-1
\end{gathered}
$$

Since -4 is the minimum value of the function, we will apply the formula for the ordinate of the vertex:

$$
\frac{4 a c-b^{2}}{4 a}=-4
$$

The solutions to the obtained system of three equations with three unknowns are $a=2, b=4, c=-2$.

The desired quadratic function is $f(x)=2 x^{2}+4 x-2$.