## Task A-2.2.

Determine, if they exist, the rational numbers $a$ and $b$ such that one solution of the quadratic equation $x^{2} + a x + b = 0$ is $\sqrt{3 + \sqrt{8}}$.
## First solution.

Let $x_{1}=\sqrt{3+\sqrt{8}}$ be a solution to the quadratic equation $x^{2}+a x+b=0$.

Notice that

$$
x_{1}=\sqrt{1^{2}+2 \sqrt{2}+(\sqrt{2})^{2}}=\sqrt{(1+\sqrt{2})^{2}}=1+\sqrt{2}
$$

Comparing with the formula for the solutions of the quadratic equation, we get

$$
\frac{-a+ \pm \sqrt{a^{2}-4 b}}{2}=x_{1}=1+\sqrt{2}=\frac{2+\sqrt{8}}{2}
$$

Since $a$ and $b$ are rational, the expression $a^{2}-4 b$ is also rational, so we conclude that the first expression in the above equality must have the + sign, and that

$$
-a=2 \quad \text { and } \quad a^{2}-4 b=8
$$

From the first equality, we conclude that $a=-2$, and from the second, $b=-1$, which are the sought rational numbers.