## Task A-2.4.

The sum of the squares of all solutions of the equation $x^{4}+a x^{2}+b=0$ is 32, and the product of all solutions of this equation is 4. Determine $a$ and $b$.
## First Solution.

Let's introduce the substitution $t=x^{2}$. This transforms the biquadratic equation into a quadratic one

$$
t^{2}+a t+b=0
$$

then from Viète's formulas, for the solutions $t_{1}$ and $t_{2}$ of this quadratic equation, we have

$$
t_{1}+t_{2}=-a \quad \text{and} \quad t_{1} \cdot t_{2}=b
$$

By reversing the substitution from $x^{2}=t_{1}$ and $x^{2}=t_{2}$, we get that the solutions of the original equation are

$$
x_{1}=\sqrt{t_{1}}, \quad x_{2}=-\sqrt{t_{1}}, \quad x_{3}=\sqrt{t_{2}} \quad \text { and } \quad x_{4}=-\sqrt{t_{2}}
$$

From the conditions of the problem, it follows that

$$
\begin{aligned}
& 4=x_{1} \cdot x_{2} \cdot x_{3} \cdot x_{4}=\left(-t_{1}\right) \cdot\left(-t_{2}\right)=b, \quad \text { i.e., } \quad b=4 \\
& 32=x_{1}^{2}+x_{2}^{2}+x_{3}^{2}+x_{4}^{2}=t_{1}+t_{1}+t_{2}+t_{2}=2\left(t_{1}+t_{2}\right)=-2 a, \quad \text { i.e., } \quad a=-16 . \quad 2 \text { points }
\end{aligned}
$$