## Task A-2.2.

Let $a, b \in \mathbb{R}$. The solutions to the quadratic equation $a x^{2} + b x + 1 = 0$ are real. If each of these solutions is decreased by 1, we get the solutions to the quadratic equation $b x^{2} + x + a = 0$.

Determine all such real numbers $a, b$.
## First Solution.

Let $x_{1}$ and $x_{2}$ be the solutions of the quadratic equation $b x^{2}+x+a=0$.

From Viète's formulas, we have

$$
x_{1}+x_{2}=-\frac{1}{b} \quad \text { and } \quad x_{1} x_{2}=\frac{a}{b}
$$

From the condition of the problem, we know that $x_{1}+1$ and $x_{2}+1$ are the solutions of the quadratic equation $a x^{2}+b x+1=0$, so again from Viète's formulas, we have

$$
-\frac{b}{a}=\left(x_{1}+1\right)+\left(x_{2}+1\right)=\left(x_{1}+x_{2}\right)+2=-\frac{1}{b}+2
$$

and

$$
\frac{1}{a}=\left(x_{1}+1\right)\left(x_{2}+1\right)=x_{1} x_{2}+\left(x_{1}+x_{2}\right)+1=\frac{a}{b}-\frac{1}{b}+1
$$

From the above equalities, after rearranging, we get the system of equations

$$
\begin{array}{r}
2 a b+b^{2}-a=0 \\
a^{2}-a+a b-b=0
\end{array}
$$

1 point

The second equation can be factored as $(a+b)(a-1)=0$, from which it follows that $a=-b$ or $a=1$.

If $a=-b$, substituting into the first equation gives $-a^{2}-a=-a(a+1)=0$, from which it follows that $a=0$ or $a=-1$. The case $a=0$ is discarded because then one of the quadratic equations in the problem statement would have a leading coefficient of zero, so we are left with $(a, b)=(-1,1)$.

If $a=1$, substituting into the second equation gives that $b$ is a solution of the quadratic equation $b^{2}+2 b-1=0$. Thus, $b=-1 \pm \sqrt{2}$, so the possible pairs are $(a, b)=(1,-1-\sqrt{2})$ and $(a, b)=(1,-1+\sqrt{2})$.

It remains to check the condition that the quadratic equations have real solutions. For this to be satisfied, their discriminants must be non-negative, i.e., $b^{2}-4 a \geqslant 0$ and $1-4 a b \geqslant 0$ must hold.

We see that all the found pairs $(a, b)$ satisfy this except for $(a, b)=(1,-1+\sqrt{2})$. 1 point Finally, the pairs $(a, b)$ that satisfy the conditions of the problem are $(-1,1)$ and $(1,-1-\sqrt{2})$.