α) Είναι:

$$$$\begin{array}{rl}A+B& ={\displaystyle \frac{1}{3-\sqrt{7}}}+{\displaystyle \frac{1}{3+\sqrt{7}}}\\ & ={\displaystyle \frac{3+\sqrt{7}}{(3+\sqrt{7})(3-\sqrt{7})}}+{\displaystyle \frac{3-\sqrt{7}}{(3+\sqrt{7})(3-\sqrt{7})}}\\ & ={\displaystyle \frac{3-\sqrt{7}+3+\sqrt{7}}{(3+\sqrt{7})(3-\sqrt{7})}}\\ & ={\displaystyle \frac{6}{3^2-(\sqrt{7}{)}^2}}\\ & ={\displaystyle \frac{6}{9-7}}\\ & ={\displaystyle \frac{6}{2}}\\ & =3\end{array}$$

Ισχύει ότι:

$$\begin{array}{rl}A\cdot B& ={\displaystyle \frac{1}{3-\sqrt{7}}}\cdot {\displaystyle \frac{1}{3+\sqrt{7}}}\\ & ={\displaystyle \frac{1}{(3+\sqrt{7})(3-\sqrt{7})}}\\ & ={\displaystyle \frac{1}{3^2-(\sqrt{7}{)}^2}}\\ & ={\displaystyle \frac{1}{9-7}}\\ & ={\displaystyle \frac{1}{2}}\end{array}$$

β) Η ζητούμενη εξίσωση είναι της μορφής:

$$x^2-Sx+P=0$$

με

$$S=A+B=3\text{ και }P={\rm A}\cdot B={\displaystyle \frac{1}{2}}$$

Τελικά μια ζητούμενη εξίσωση είναι η:

$$x^2-3x+{\displaystyle \frac{1}{2}}=0$$